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#### Answer:

According to the right-hand thumb rule, if the thumb of our right hand points in the direction of the current flowing, then the curling of the fingers will show the direction of the magnetic field developed due to it and vice versa. Let us consider the case where an electric current flows north to south in a wire.
According to the right-hand thumb rule,
(a) for any point in the east of the wire, the magnetic field will come out of the plane of paper
(b) for a point in the west of the wire, the magnetic field will enter the plane of paper
(c) for any point vertically above the wire, the magnetic field will be from right to left
(d) for any point vertically below the wire, the magnetic field will be from left to right

#### Answer:

The magnetic field due to a long, straight wire is given by

(In terms of ε0ci and d)

#### Answer:

According to the right-hand thumb rule, if we curl the fingers of our right hand in the direction of the current flowing, then the thumb will point in the direction of the magnetic field developed due to it and vice versa. Therefore, in this case, the field at the centre is going away from us.

#### Answer:

In Ampere's law i is the total current crossing the area bounded by the closed curve. The magnetic field B on the left-hand side is the resultant field due to all existing currents.

#### Answer:

The magnetic field due to a long solenoid is given as B = µ0ni, where n is the number of loops per unit length. So, if we add more loops at the ends of the solenoid, there will be an increase in the number of loops and an increase in the length, due to which the ratio n will remain unvaried, thereby leading to not a considerable effect on the field inside the solenoid.

#### Answer:

Ampere's law is valid for a loop that is not circular. However, it should have some charge distribution in the area enclosed so as to have a constant electric field in the region and a non-zero magnetic field. Even if the loop defined does not have its own charge distribution but has electric influence of some other charge distribution, it can have some constant magnetic field ().

#### Answer:

The magnetic force on a wire carrying an electric current i is $\stackrel{\to }{F}=i.\left(\stackrel{\to }{l}×\stackrel{\to }{B}\right)$, where l is the length of the wire and B is the magnetic field acting on it. If a uniformly charged ring starts rotating around a straight wire, then according to the right-hand thumb rule, the magnetic field due to the ring on the current carrying straight wire placed at its axis will be parallel to it. So, the cross product will be
$\left(\stackrel{\to }{l}×\stackrel{\to }{B}\right)=0\phantom{\rule{0ex}{0ex}}⇒\stackrel{\to }{F}=0$
Therefore, no magnetic force will act on the wire.

#### Answer:

The magnetic force on a wire carrying an electric current i is $\stackrel{\to }{F}=i.\left(\stackrel{\to }{l}×\stackrel{\to }{B}\right)$, ​where l is the length of the wire and B is the magnetic field acting on it. Suppose we have one wire in the horizontal direction (fixed) and other wire in the vertical direction (free to move). If the horizontal wire is carrying current from right to left is held fixed perpendicular to the vertical wire, which is free to move, the upper portion of the free wire will tend to move in the left direction and the lower portion of the wire will tend to move in the right direction, according to Fleming's left hand rule, as the magnetic field acting on the wire due to the fixed wire will point into the plane of paper above the wire and come out of the paper below the horizontal wire and the current will flow in upward direction in the free wire. Thus, the free wire will tend to become parallel to the fixed wire so as to experience maximum attractive force.

#### Answer:

Two proton beams going in the same direction repel each other, as they are like charges and we know that like charges repel each other.
When a charge is in motion then a magnetic field is associated with it. Two wires carrying currents in the same direction produce their fields (acting on each other) in opposite directions so the resulting magnetic force acting on them is attractive. Due to the magnetic force, these two wires attract each other.
But when a charge is at rest then only an electric field is associated with it and no magnetic fiels is produced by it. So at rest, it repels a like charge by exerting a electric force on it.
Charge in motion can produce both electric field and magnetic field.
The attractive force between two current carrying wires is due to the magnetic field and repulsive force is due to the electric field.

#### Answer:

We can obtain a magnetic field due to a straight, long wire using Ampere's law by mentioning the current flowing in the wire, without emphasising on the source of the current in the wire. To apply Ampere's circuital law, we need to have a constant current flowing in the wire, irrespective of its source.

#### Answer:

Connecting wires carrying currents in opposite directions are twisted together in using electrical appliances.If the wire is twisted, then the resultant fields from consecutive twists are in opposite directions. So the cumulative effect over a long length of wire is roughly zero.

Let at any point in between the two wires, B1 and B2 be the magnetic field due to wire 1 and wire 2 respectively.

From the diagram, we can see that the net magnetic field due to first turn is into the paper and due to second twisted turn is out of the plane of paper so these fields will cancel each other. Hence if the wire is twisted, then the resultant fields from consecutive twists are in opposite directions. So the cumulative effect over a long length of wire is roughly zero.

#### Answer:

Magnetic field can not do any work and hence can never speed up or down a particle.
Consider 2 wires carrying current in upward direction.
Magnetic field due to current in wire 1 produces a magnetic field out of the plane of paper at the position of wire 2. Due to this magnetic field, a force is exerted on wire 2. Wire 2 electron, moving in downward direction,  move in circular paths due to this magnetic force. As these electrons can not come out of the wire so while describing circular path,they hit the edges of the wire and tranfer a momentum to the wire. Due to this change in momentum, wire starts moving and gains kinetic energy.

#### Answer:

(c) upwards

A vertical wire is carrying current in upward direction, so the magnetic field produced will be anticlockwise (according to the right-hand thumb rule). As the electron beam is sent horizontally towards the wire, the direction of the current will be horizontally away from the wire (direction of conventional current is opposite to the direction of the negative charge). According to Fleming's left-hand rule, the force will act in upward direction, deflecting the beam in the same direction.

#### Answer:

(c) will not exert any force on the circular loop

The magnetic force on a wire carrying an electric current i is given as​ $\stackrel{\to }{F}=i.\left(\stackrel{\to }{l}×\stackrel{\to }{B}\right)$, where l is the length of the wire and B is the magnetic field acting on it. If a current-carrying straight wire is kept along the axis of a circular loop carrying current, then according to the right-hand thumb rule, the magnetic field due to the wire on the current-carrying loop will be along its circumference, which contains a current element .
So, the cross product will be​
$\left(\stackrel{\to }{l}×\stackrel{\to }{B}\right)=0\phantom{\rule{0ex}{0ex}}⇒\stackrel{\to }{F}=0$
Thus, the straight wire will not exert any force on the loop.

#### Answer:

(a) towards the proton beam
A proton beam is going from north to south, so the direction of the current due to the beam will also be from north to south. Also, an electron beam is going from south to north, so the direction of the current due to the beam will also be  from north to south. The direction of conventional current is along the direction of the flow of the positive charge and opposite to the flow of the negative charge. The magnetic field generated due to them will enter the plane of paper in the west and come out of the plane of paper in the east, according to the right-hand thumb rule. Since both the beams have currents in the same direction, they will apply equal and opposite forces on each other and, hence, will attract each other. Thus, the electron beam will be deflected towards the proton beam.

#### Answer:

(d) is towards west at both A and B
According to the right-hand thumb rule, if we curl the fingers of our right hand in the direction of the current flowing, then the stretching of the thumb will show the direction of the magnetic field developed due to it and vice versa.
Let north-south is along x axis and east-west is along y axis. Circular wire is in xz plane. Then point A will lie on positive y axis and B on negative y axis. On looking from point B, current is flowing in anticlockwise direction so the magnetic field will point from right to left. Hence, the magnetic field due to the loop will be towards west at both A and B.

#### Answer:

(b) move towards the wire

Force acting on the wire per unit length carrying current i2 due to the wire carrying current i1 placed at a distance d is given by

So, forces per unit length acting on sides AB and CD are as follows:

Here, FAB > FCD  because force is inversly proportional to the distance from the wire and wire AB is closer to the wire carrying current i1.
The forces per unit length acting on sides BC and DA will be equal and opposite, as they are equally away from the wire carrying current i1, with current i2 flowing in the opposite direction.
FBC $-$FDA
Now,
Net force:

(Towards the wire)
Therefore, the loop will move towards the wire.

#### Answer:

(d) zero

The force on a charged particle q moving with velocity v in a magnetic field B is given by
$\stackrel{\to }{F}=q\left(\stackrel{\to }{v}×\stackrel{\to }{B}\right)$
As the charge is moving along the magnetic line of force, the velocity and magnetic field vectors will point in the same direction, making a cross product.

So, the magnetic force on the particle will be zero.

#### Answer:

(c) both of them

Because of the presence of a charge, a particle produces an electric field. Also, because of its motion, that is, the flow of charge or current, there is generation of a magnetic field.

#### Answer:

(c) the kinetic energy

When a particle of mass m carrying charge q is projected with speed v in a plane perpendicular to a uniform magnetic field B, the field tends to deflect the particle in a circular path of radius r

Kinetic energy of the particle,
Therefore, the area bounded is proportional to the kinetic energy.

#### Answer:

(c)$\frac{{{R}_{1}}^{2}}{{{R}_{2}}^{2}}$
Particles X and Y of respective masses m1 and m2 are carrying charge q describing circular paths with respective radii R1 and R2 such that

Since both the particles are accelerated through the same potential difference, both will have the same kinetic energy.

#### Answer:

(b) towards 40 A

​According to Fleming's left-hand rule, if the forefinger and middle finger of our left hand point towards the magnetic field acting on a wire and the current flowing in the wire, respectively, then the thumb will point towards the direction in which the force will act (keeping all three perpendicular). Direction of force can be determined using Fleming's left-hand rule.

In the figure, dotted circle shows the magnetic filed lines due to both the wires.
Magnetic field at any point on the middle wire will be acting along the tangent to the masgnetic field lines at that point.
Therefore, the wire will experience a magnetic field pointing ​towards the 40 A wire.
Due to AB, the force will be towards right and due to CD, the force on the wire will be towards right. So, both the forces will add to give a resultant force, which will be towards right, that is, towards the 40 A current-carrying wire.

#### Answer:

(c) 2

The magnetic field due to the current-carrying long, straight wire at point a is given by

When both the wires carry currents i1 and i2 in the same direction, they produce magnetic fields in opposite directions at any point in between the wires.

Here, a is the distance of the midpoint from both the wires.
When both the wires carry currents in opposite directions, they produce fields in the same direction at the midpoint of the two wires.

On solving eqs. (1) and (2), we get

#### Answer:

(a) $\frac{{\mathrm{\mu }}_{o}i}{2\mathrm{\pi }r}$
Magnetic field will be independent of the motion of the observer because the velocity with which the observer is moving is comparable to drift velocity of electron which is very small as compared to the speed of flow of current from one end of wire to other end. So it can be neglected  and hence, magnetic field due to the wire w.r.t the observer will be
B = $\frac{{\mathrm{\mu }}_{o}i}{2\mathrm{\pi }r}$

μ0i2πr

#### Answer:

(a)
(b)

The magnetic field at the origin due to current element  placed at a position $\stackrel{\to }{r}$ is given by

According to the cross product property,

#### Answer:

(a) x, y have the same dimensions.
(b) y, z have the same dimensions.
(c) z, x have the same dimensions.

Lorentz Force:

Time constant of RC circuit = RC so dimensionally [RC] = [T]

Therefore, x, y and z have the same dimensions.

#### Answer:

(b) the directions of the magnetic fields are the same
(c) the magnitudes of the magnetic fields are equal
(d) the field at one point is opposite to that at the other point.
Consider a current carrying wire lying along x axis.
At any two points on z axis which are at equal distance from the wire,one above the wire and one below the wire,the magnitude of magnetic field will be same and their directions will be opposite to each other.
At any two points on z axis which are at different distances from the wire,one above the wire and other also above the wire,the magnitude of magnetic field will be different and their directions will be same to each other.

#### Answer:

(b) minimum at the axis of the wire
(c) maximum at the surface of the wire

A long, straight wire of radius R is carrying current i, which is uniformly distributed over its cross section. According to Ampere's law,

Here i, is the current enclosed by the amperian loop drawn inside the wire.
Binside will be proportional to the distance from the axis.
On the axis
B =0
The magnetic fields from points on the cross section will point in opposite directions and will cancel each other at the centre.

Therefore, the magnitude of the magnetic field is maximum at the surface of the wire and minimum at the axis of the wire.

#### Answer:

(b) is constant inside the tube
(c) is zero at the axis

A hollow tube is carrying uniform electric current along its length, so the current enclosed inside the tube is zero.
According to Ampere's law,

The  magnetic fields from points on the circular surface will point in opposite directions and cancel each other.

#### Answer:

(a) outside the cable
(b) inside the inner conductor
According to Ampere's law, in a coaxial, straight cable carrying currents i in the inner conductor and -i (equally in the opposite direction) in the outside conductor.
Inside the inner conductor

In between the 2 conductors

Outside the outer conductor

Therefore, the magnetic field is zero outside the cable.

#### Answer:

(b) The magnetic field at the axis of the conductor is zero.
(c) The electric field in the vicinity of the conductor is zero.
As the current is flowing through a conductor so it it is distributed only on the surface of the conductor not in the volume of the cylindrical conductor. It is equivalent to charge distribution on a cylindrical sheet for which electric field inside a conducting cylindrical sheet is always zero.
Magnetic fields at any point inside the conducting cylinder is proportional to the distance from the axis of the cylinder.
At the axis, r = 0. This implies that field will be zero at the axis.

#### Answer:

Using the relation , we get

Units of
Force (F) = N
Current (I) = A
Time (T) = s
Velocity (v) = m/s
$⇒B=\mathrm{N}/\mathrm{A}-\mathrm{m}$
Now, using the relation $B=\frac{{\mu }_{0}i}{2\mathrm{\pi }r}$, we get
$⇒{\mu }_{0}=B\frac{2\mathrm{\pi }r}{i}=\frac{\mathrm{N}}{\mathrm{A}-\mathrm{m}}×\frac{\mathrm{m}}{\mathrm{A}}\phantom{\rule{0ex}{0ex}}⇒{\mu }_{0}\mathit{=}\mathrm{N}/{\mathrm{A}}^{2}$

#### Answer:

Given:
Magnitude of current, I = 10 A
Separation of the point from the wire, d = 1 m

The magnetic field  at point (1 m, 0, 0) is given by

(Along the +ve y-axis by the right-hand thumb rule)

#### Answer:

Given:
Magnitude of current, I = 10 A
Diameter of the wire, d = 1.6 × 10−3 m
∴ Radius of the wire = 0.8 × 10−3 m

The magnetic field intensity is given by
$\mathrm{B}=\frac{{\mu }_{0}i}{2\mathrm{\pi }r}$

#### Answer:

Given:
Magnitude of current, I = 100 A
Separation of the road from the wire, d = 8 m
Thus, the magnetic field is given by

#### Answer:

Given:
Uniform magnetic field, B0 = 1.0 × 10−5 T    (Vertically upwards)
Separation of the point from the wire, d = 2 cm = 0.02 m
The magnetic field due to the wire is given by

Now,
Net magnetic field at point P:
BPBw + B0 = 2 × 10−5 T
Net magnetic field at point Q:
BQBw B0 = 0

#### Answer:

(a) As the wire in question is carrying current, so it will also generate a magnetic field around it. And for a long straight wire it will be maximum at the mid-point called P.
Now,
Magnetic field generated by the current carrying wire$=\frac{{\mathrm{\mu }}_{\mathrm{o}}i}{2\pi r}$
Net magnetic field = $\mathrm{B}+\frac{{\mathrm{\mu }}_{0}i}{2\mathrm{\pi }r}$

(b) Magnetic field B = 0
when $r<\frac{{\mathrm{\mu }}_{0}i}{2\mathrm{\pi B}}$
Clearly,
B = 0
when $r=\frac{{\mathrm{\mu }}_{0}i}{2\mathrm{\pi B}}$
But when $r>\frac{{\mathrm{\mu }}_{0}i}{2\mathrm{\pi B}}$,
Net magnetic field = $\mathrm{B}-\frac{{\mathrm{\mu }}_{0}i}{2\mathrm{\pi }r}$

#### Answer:

Given:
Uniform magnetic field, B0 = 4.0 × 10−4 T
Magnitude of current, I = 30 A
Separation of the point from the wire, d = 0.02 m
Thus, the magnetic field due to current in the wire is given by

B0 is perpendicular to B (as shown in the figure).
∴ Resultant magnetic field

#### Answer:

Given:
Uniform magnetic field, B0 = 2.0 × 10−3 T (From south to north)
To make the resultant magnetic field zero, the magnetic field due to the wire should be of the same magnitude as B0 and in the direction north to south.
The above condition will be satisfied when the required point will be placed in the west w.r.t. the wire.
Let the separation of the point from the wire be d.

The magnetic field due to current in the wire is given by
$B=\frac{{\mu }_{0}I}{2\pi d}$

From the question, B = B0.

#### Answer:

For point A1,
Magnitude of current in wires, I = 10 A
Separation of point A1 from the wire on the left side, d = 2 cm
Separation of point A1 from the wire on the right side, d' = 6 cm

In the figure
Red and blue arrow denotes the direction of magnetic field due to the wire marked as red and blue respectively.
P (marked red) denotes the wire carrying current in a plane going into the paper.
Q (marked blue) denotes the wire carrying current in a plane coming out of the paper.
Also from the figure, we can see that

The magnetic field at A1 due to current in the wires is given by
$B=\frac{{\mu }_{0}I}{2\mathrm{\pi }d}-\frac{{\mu }_{0}I}{2\mathrm{\pi }d\text{'}}$    ...(1)

Similarly, we get the magnetic field at A2 using eq. (1).

Now,
Magnetic field at A3:

Magnetic field at A4:
Separation of point A4 from the wire on the left side, d =
Separation of point A4 from the wire on the right side, d' =
Thus, the magnetic field at A4 due to current in the wires is given by

#### Answer:

Given:
Magnitude of currents, I1 = I2 = 10 A
Separation of the point from the wires, d = 2 cm
Thus, the magnetic field due to current in the wire is given by
${B}_{1}={B}_{2}=\frac{{\mu }_{0}I}{2\mathrm{\pi }d}$

In the figure, dotted circle shows the magnetic field lines due to current carrying wire placed in a plane perpendicular to the plane of the paper.
From the figure, we can see that is an equilateral triangle.

Angle between the magnetic fields due to current in the wire, θ = $60°$
∴ Required magnetic field at P
${B}_{\mathrm{net}}=\sqrt{{{B}_{1}}^{2}+{{B}_{2}}^{2}+2{B}_{1}{B}_{2}\mathrm{cos}\theta }$

#### Answer:

Given:
Magnitude of current, I = 5 A
Separation of the point from the wire, d = 1 m
Thus, the magnitude of magnetic field due to current in the wires is given by
${B}_{1}={B}_{2}=\frac{{\mu }_{0}I}{2\pi d}$

(a) At point (1 m, 1 m), the magnetic fields due to the wires are the same in magnitude, but they are opposite in direction.
Hence, the net magnetic field is zero.

(b) At point (−1 m, 1 m), the magnetic fields due to the wires are in upward direction.

= 2 × 10−6 T      (Along the z-axis)

(c) At point (−1 m, −1 m), the magnetic fields due to the wires are the same in magnitude, but they are opposite in direction.
Hence, the net magnetic field is zero.

(d) At point (1 m, −1 m), the magnetic fields due to the wires are in upward direction.

= 2 × 10−6 T    (Along the negative z-axis)

#### Answer:

Given:

Let the horizontal wires placed at the bottom and top are denoted as W1 and W2 respectively.
Let the vertical wires placed at the right and left to point P are denoted as W3 and W4 respectively.
Magnitude of current, I = 5 A
(a) Consider point P.
Magnetic fields due to wires W1 and W2 are the same in magnitude, but they are opposite in direction.
Magnetic fields due to wires W3 and W4 are the same in magnitude, but they are opposite in direction.
Hence, the net magnetic field is zero.
Net magnetic field at P due to these four wires = 0

(b) Consider point Q1.
Due to wire W1, separation of point Q1 from the wire (d) is 7.5 cm.
So, the magnetic field due to current in the wire is given by
${B}_{{W}_{1}}=\frac{{\mu }_{0}I}{2\pi d}$
= 4 × 10−5 T     (In upward direction)

Due to wire W2, separation of point Q1 from the wire (d) is 2.5 cm.
So, the magnetic field due to current in the wire is given by
(In upward direction)

Due to wire W3, separation of point Q1 from the wire (d) is 7.5 cm.
So, the magnetic field due to current in the wire is given by
BW3 = 4 × 10−5 T   (In upward direction)

Due to wire W4, separation of point Q1 from the wire (d) is 2.5 cm.
So, the magnetic field due to current in the wire is given by
(In upward direction)

∴ Net magnetic field at point Q1

At point Q2,
Magnetic field due to wire W1:
BW1 = 4 × 10−5 T   (In upward direction)
Magnetic field due to wire W2:
(In upward direction)
Magnetic field due to wire W3:
(In downward direction)
Magnetic field due to wire W4:
(In downward direction)

∴ Net magnetic field at point Q2,
${B}_{{Q}_{2}}=0$

Similarly, at point Q3, the magnetic field is 1.1 × 10−4 T (in downward direction) and at point Q4, the magnetic field is zero.

#### Answer:

As shown in the figure, points P, Q, R and S lie on a circle of radius d.
Let the wires be named W1 and W2.

Now,
At point P, the magnetic field due to wire W1 is given by
B1 = 0
At point P, the magnetic field due to wire W2 is given by
(Perpendicular to the plane in outward direction)
$⇒{B}_{\mathrm{net}}=\frac{{\mathrm{\mu }}_{0}i}{4\mathrm{\pi }d}$ (Perpendicular to the plane in outward direction)

At point Q, the magnetic field due to wire W1 is given by
(Perpendicular to the plane in inward direction)
At point Q, the magnetic field due to wire W2 is given by
B2 = 0
$⇒{B}_{\mathrm{net}}=\frac{{\mathrm{\mu }}_{0}i}{4\mathrm{\pi }d}$ (Perpendicular to the plane in inward direction)

At point R, the magnetic field due to wire W1 is given by
B1 = 0
At point R, the magnetic field due to wire W2 is given by
(Perpendicular to the plane in inward direction)
$⇒{B}_{\mathrm{net}}=\frac{{\mathrm{\mu }}_{0}i}{4\mathrm{\pi }d}$ (Perpendicular to the plane in inward direction)

At point S, the magnetic field due to wire W1 is given by
(Perpendicular to the plane in outward direction)
At point S, the magnetic field due to wire W2 is given by
B2 = 0
$⇒{B}_{\mathrm{net}}=\frac{{\mathrm{\mu }}_{0}i}{4\mathrm{\pi }d}$ (Perpendicular to the plane in outward direction)
Hence, the magnitude of the magnetic field at points P, Q,  R and S is $\frac{{\mathrm{\mu }}_{0}i}{4\mathrm{\pi }d}$.

#### Answer:

Let AB be the wire of length x with midpoint O.
Given:
Magnitude of current = i
Separation of the point from the wire = d

Now,
The magnetic field on a perpendicular bisector is given by

So, if  d > > x (neglecting x), then
$B=\frac{{\mathrm{\mu }}_{0}i}{4\mathrm{\pi }d}\frac{2x}{2d}\phantom{\rule{0ex}{0ex}}⇒B\propto \frac{1}{{d}^{2}}$
And, if d < < x (neglecting d), then

#### Answer:

Let AB be the wire of length 10 cm and P be the required point.
Given:
Magnitude of current, i = 10 A
The angles made by points A and B with point P are , respectively.
∴ Separation of the point from the wire, d

Thus, the magnetic field due to current in the wire is given by

#### Answer:

Given:
Magnitude of current = i
Separation of the point from the wire = d

Thus, the magnetic field due to current in the long wire is given by
${B}_{1}=\frac{{\mu }_{0}i}{2\mathrm{\pi }d}\phantom{\rule{0ex}{0ex}}$
Also, the magnetic field due to a section of length l on a perpendicular bisector is given by
${B}_{2}=\frac{{\mu }_{0}i}{4\pi d}\frac{2l}{\sqrt{{l}^{2}+4{d}^{2}}}$

Now,
B1 > B2
According to the question,

#### Answer:

Let the currents in wires ABC and ADC be i1 and i2, respectively.
The resistances in wires ABC and ADC are r and 2r, respectively.

And,

Using (1) and (2), we get
and
The angles made by points A and D with point O are , respectively.
Separation of the point from the wire, d = a/2
Now,
The magnetic field due to current in wire AD is given by

The magnetic field at centre due to wire ADC is given by

(Perpendicular to the plane in outward direction)

The magnetic field at centre due to wire ABC is given by

(Perpendicular to the plane in inward direction)

(Perpendicular to the plane in inward direction)

#### Answer:

B at P due to AD = $\frac{{\mathrm{\mu }}_{0}}{4\pi }.\frac{i}{2}.\frac{4}{{d}^{2}}.a\left[\frac{\left(\frac{a}{2}\right)}{\sqrt{{\left(\frac{a}{2}\right)}^{2}+{\left(\frac{a}{4}\right)}^{2}}}+\frac{\left(\frac{a}{2}\right)}{\sqrt{{\left(\frac{a}{2}\right)}^{2}+{\left(\frac{3a}{4}\right)}^{2}}}\right]$along•
$\frac{{\mathrm{\mu }}_{0}i}{4\pi a}\left[\frac{\left(\frac{a}{2}\right)}{\sqrt{{\left(\frac{a}{2}\right)}^{2}+{\left(\frac{a}{4}\right)}^{2}}}+\frac{\left(\frac{a}{2}\right)}{\sqrt{{\left(\frac{a}{2}\right)}^{2}+{\left(\frac{3a}{4}\right)}^{2}}}\right]$along•

B at P due to AC=$\frac{{\mathrm{\mu }}_{0}}{4\pi }.\frac{i}{2}.\frac{16}{9{a}^{2}}.a.2\left[\frac{\left(\frac{3a}{4}\right)}{\sqrt{{\left(\frac{3a}{4}\right)}^{2}+{\left(\frac{a}{2}\right)}^{2}}}\right]$
$=\frac{4{\mathrm{\mu }}_{0}i}{9\pi a}\left[\frac{\left(\frac{3a}{4}\right)}{\sqrt{{\left(\frac{a}{4}\right)}^{2}+{\left(\frac{3a}{2}\right)}^{2}}}\right]$along•
B at P due to AB = $\frac{{\mathrm{\mu }}_{0}}{4\pi }.\frac{i}{2}.\frac{16}{9{a}^{2}}.a.2\left[\frac{\left(\frac{3a}{4}\right)}{\sqrt{{\left(\frac{3a}{4}\right)}^{2}+{\left(\frac{a}{2}\right)}^{2}}}\right]$along•
B at P due to BC = $\frac{{\mathrm{\mu }}_{0}}{4\pi }.\frac{i}{2}.\frac{4}{{a}^{2}}.a.\left[\frac{\left(\frac{a}{2}\right)}{\sqrt{{\left(\frac{a}{2}\right)}^{2}+{\left(\frac{a}{4}\right)}^{2}}}+\frac{\left(\frac{a}{2}\right)}{\sqrt{{\left(\frac{a}{2}\right)}^{2}+{\left(\frac{3a}{4}\right)}^{2}}}\right]$ along•
$=\frac{{\mathrm{\mu }}_{0}i}{2\pi a}\left[\frac{\left(\frac{a}{2}\right)}{\sqrt{{\left(\frac{a}{2}\right)}^{2}+{\left(\frac{a}{4}\right)}^{2}}}+\frac{\left(\frac{a}{2}\right)}{\sqrt{{\left(\frac{a}{2}\right)}^{2}+{\left(\frac{3a}{4}\right)}^{2}}}\right]\mathrm{along}\otimes$
So, net magnetic field at point P.
$\mathrm{B}=\frac{4{\mathrm{\mu }}_{0}i}{\mathrm{\pi }a}\left[\frac{\left(\frac{\mathrm{a}}{4}\right)}{\sqrt{{\left(\frac{\mathrm{a}}{2}\right)}^{2}+{\left(\frac{\mathrm{a}}{4}\right)}^{2}}}\right]-\frac{4{\mathrm{\mu }}_{0}i}{9\mathrm{\pi }a}\left[\frac{\left(\frac{3\mathrm{a}}{4}\right)}{\sqrt{{\left(\frac{\mathrm{a}}{2}\right)}^{2}+{\left(\frac{3\mathrm{a}}{4}\right)}^{2}}}\right]\phantom{\rule{0ex}{0ex}}=\frac{4{\mathrm{\mu }}_{0}i}{\mathrm{\pi }a}\frac{1}{4}\left[\frac{1}{\sqrt{\frac{1}{4}+\frac{1}{16}}}\right]-\frac{4{\mathrm{\mu }}_{0}i}{9\mathrm{\pi }a}.\frac{3}{4}\left[\frac{1}{\sqrt{\frac{1}{4}+\frac{9}{16}}}\right]$
$=\frac{4{\mathrm{\mu }}_{0}i}{4\pi a}\left[\frac{4}{\sqrt{5}}\right]-\frac{{\mathrm{\mu }}_{0}i}{3\pi a}\left[\frac{4}{\sqrt{13}}\right]\left[\frac{4}{\sqrt{13}}\right]\phantom{\rule{0ex}{0ex}}$ along•
$=\frac{4{\mathrm{\mu }}_{0}i}{2\pi a}\left[\frac{1}{\sqrt{5}}-\frac{1}{3\sqrt{13}}\right]$ along•
$=\frac{2{\mathrm{\mu }}_{0}i}{\pi a}\left[\frac{1}{\sqrt{5}}-\frac{1}{3\sqrt{13}}\right]$ along•

#### Answer:

The loop ABCD can be considered as a circuit with two resistances in parallel, one along branch AB and other along branch ADC.

As, the sides of the loop are identical, their resistances are also same.

Let the resistance of each side be r.

The resistance of branch AB = r

The resistance of branch ADC = 3r

The current in the branches are calculated as:

${i}_{\mathrm{AB}}=i×\frac{3r}{3r+r}=\frac{3i}{4}$

${i}_{\mathrm{ADC}}=i×\frac{r}{3r+r}=\frac{i}{4}$
As current follow the least resistive path so

Current in branch AB = $\frac{3i}{4}$
Current in branch ADC = $\frac{i}{4}$
At the centre of the loop:

Magnetic field due to wire AD, DC and CB will be into the plane of paper according to right hand thumb rule.
Magnetic field due to wire AB will be out of the plane of paper according to right hand thumb rule.
Net magnetic field at the centre = BAD +BDC +BCB − BAB  which will be out of the plane of paper.
As, perpendicular distance of the centre from every wire will be equal to $\frac{a}{\sqrt{2}}$ and angle made by corner points of each side at the centre is $45°$.

#### Answer:

Let current 2I enter the circuit.
Since the wire is uniform, the current will be equally divided at point A (as shown in the figure).

Now,
Magnetic field at P due to wire AB = B (say)
(Perpendicular to the plane in outward direction)

Magnetic field at P due to wire BD = B' (say)
(Perpendicular to the plane in outward direction)

Magnetic field at P due to wire AC = Magnetic field at P due to wire AB = B
(Perpendicular to the plane in inward direction)

Magnetic field at P due to wire CD = Magnetic field at P due to wire BD = B'
(Perpendicular to the plane in inward direction)

∴ Net magnetic field at P = B + B' − B − B' = 0

#### Answer:

Let ABC be the equilateral triangle with side l/3 and centre M.
(a)

The angles made by points B and C with centre M are , respectively.
Separation of the point from the wire, d = MO = $\frac{l}{6\sqrt{3}}$
Thus, the magnetic field due to current in wire BC is given by

$⇒B=\frac{{\mathrm{\mu }}_{0}i}{4\pi l}6\sqrt{3}×\sqrt{3}$
Now,
Net magnetic field at M = Magnetic field due to wire BC + Magnetic field due to wire CA + Magnetic field due to wire AB
Since all wires are the same,
$=\frac{27{\mathrm{\mu }}_{0}i}{\mathrm{\pi l}}$
It is perpendicular to the plane in outward direction if the current is anticlockwise and perpendicular to the plane in inward direction if the current is clockwise.

(b)
The angles made by points B and C with centre M are , respectively.
Separation of the point from the wire, d = l/8
Thus, the magnetic field due to current in wire BC is given by

Since all wires are the same,
Net magnetic field at M = 4 × Magnetic field due to wire BC
$\frac{8\sqrt{2}{\mathrm{\mu }}_{0}i}{\mathrm{\pi }l}$

#### Answer:

Let CAB be the wire making an angle α, P be the point on the bisector of this angle situated at a distance x from the vertex A and d be the perpendicular distance of AC and AB from P.

From the figure,

The angles made by points A and C with point P are , respectively.
Separation of the point from the wire,
Thus, the magnetic field due to current in wire AC is given by

Now, the magnetic field due to wires AC and AB is given by

#### Answer:

Let the angles made by points A and B with point O be , respectively.
And,

Separation of the point from the wire, d = b/2

And,

Thus, the magnetic field due to current in wire AB is given by

Similarly, the magnetic field due to current in wire BC is given by

Now,
Magnetic field due to current in wire CD = Magnetic field due to current in wire AB = B
And,
Magnetic field due to current in wire DA = Magnetic field due to current in wire BC = B'

#### Answer:

(a)
Using figure,

(b)
when n→ ∞, polygon becomes a circle with radius r
and magnetic field will become
$B=\frac{{\mathrm{\mu }}_{0}i}{2r}$

#### Answer:

By appluing Kirchoff voltagee law,we can see that the current in the circuit is zero.
Net current in the circuit = 0
As
magnetic field is always proportional to the current flowing in the circuit hence,
Net magnetic field at point P = 0
Field at point P is independent of the values of the resistances in the circuit.

#### Answer:

Given:
Magnitude of current in both wires, i1 = i2 = 20 A
Force acting on 0.1 m of the second wire, F = 2.0 × 10−5 N
∴ Force per unit length = $\frac{2×{10}^{-5}}{0.1}$ = 2.0 × 10−4 N/m
Now,
Let the separation between the two wires be d.
Thus, the force per unit length is given by

#### Answer:

Let wires W1, W2 and W3 be arranged as shown in the figure.

Given:
Magnitude of current in each wire, i1 = i2 = i3 = 10 A
The magnetic force per unit length on a wire due to a parallel current-carrying wire is given by

So, for wire W1,

For wire W2,

For wire W3,

= 6 × 10−4 N

#### Answer:

Let the third wire W3 having current i in upward direction be placed x cm from the 10 A current wire.

The magnetic force per unit length on a wire due to a parallel current-carrying wire is given by

According to the question, wire W3 experiences no magnetic force.
∴

Thus, wire W3 is placed 2 cm from the 10 A current wire.

#### Answer:

Since wires AB, CD and EF have identical resistance, the current (30 A) gets equally distributed in them, that is, 10 A in each wire.

The magnetic force per unit length on a wire due to a parallel current-carrying wire is given by

∴ Magnetic force per unit length of AB = Force due to current in CD + Force due to current in EF

Similarly,
Magnetic force per unit length of CD = Force due to current in AB − Force due to current in EF
∵ Force on CD due to current in AB = Force due to current in EF
∴ Magnetic force per unit length of CD = 0

#### Answer:

Given:
Magnitude of current, i1 = 10 A
Separation between two wires, d = 5 mm
Linear mass density of the second wire, λ = 1.0 × 10−4 kgm−1
Now,
Let i2 be the current in the second wire in opposite direction.
Thus, the magnetic force per unit length on the wire due to a parallel current-carrying wire is given by
(upwards)
Also,
Weight of the second wire, W = mg
Weight per unit length of the second wire, (downwards)

Now, according to the question,

#### Answer:

Given:
Current in the loop, i1 = 6 A
Current in the wire, i2 = 10 A

Now, consider an element on PQ of width dx at a distance x from the wire.

Force on the element is given by
$dF=\frac{{\mu }_{0}{i}_{1}{i}_{2}}{2\pi x}dx\phantom{\rule{0ex}{0ex}}$
Force acting on part PQ is given by

Both forces are equal in magnitude, but they are opposite in direction.

(b) The magnetic field intensity due to wire on SP is given by
$B=\frac{{\mu }_{0}{i}_{2}}{2\mathrm{\pi }r}$

Force on part SP is given by

(Towards right)

Force on part RQ is given by

(Towards left)

Thus, the net force on the loop is given by

(Towards right)

#### Answer:

Given:
No. of turns, n = 1
Magnitude of current, i = 5.00 A
Now, let the radius of the loop be r.
Thus, the magnetic field at the centre due to the current in the loop is given by

#### Answer:

Given:
No. of turns, n = 100
Radius of the loop, r = 5 cm = 0.05 m
Magnetic field intensity, B = 6.0 × 10−5 T
Now, let the magnitude of current be i.

#### Answer:

Given:
Frequency of the electron = 3 × 105
Time taken by the electron to complete one revolution, T = $\frac{1}{\mathrm{Frequency}}$
Current in the circle, i = $\frac{q}{t}$
Radius of the loop, r = 0.5 ${\mathrm{A}}^{°}$ =
Thus, the magnetic field at the centre due to the current in the loop is given by

#### Answer:

As the centre of the loop, that is, point O, lies on the same line of two long, straight wires, the magnetic field at O due to each straight wire is zero.

Since wires ABC and ADC are identical, the current gets equally distributed in two parts at point A. So, the magnetic field due to ABC and ADC at O are equal in magnitude but are opposite in directions. (as shown in the figure).
∴ Net magnetic field at O = 0

#### Answer:

No. of turns: n1 = 50 and n2 = 100
Magnitude of currents: i1 = i2 = 2 A
Radii of loops: r1 = 5 cm and r2 = 10 cm
(a) In the same sense:

The magnetic field intensity at the centre is given by

(b) In the opposite sense:
The magnetic field intensity at the centre is given by

#### Answer:

Given:
No. of turns: n1 = 50 and n2 = 100
Magnitude of currents: i1 = i2 = 2 A
Radii of loops: r1 = 5 cm and r2 = 10 cm
(a) In the same sense:

The magnetic field intensity at the centre due to C1 is given by

(In the plane of paper in upward direction)

The magnetic field intensity at the centre due to C2 is given by

(In the plane of paper in upward direction)

In this case, magnetic fields due to C1 and C2 at the centre are along the same direction.
Thus, the net magnetic field is given by

(b) When the direction of current in the two coils is opposite to each other then the magnetic fields will also point in opposite directions as shown in the figure. Hence, the net magnetic field will be obtained by the subtraction of the two magnetic fields.

#### Answer:

Given:
Magnitude of current in the loop, I = 10 A
Radius of the loop, r = 20 cm = 20 × 10−2 m
Thus, the magnetic field intensity at the centre is given by

Now,
Velocity of the electron, v = 2 × 106 m/s
Angle between the velocity and the magnetic field intensity, θ = 30°
Thus, the magnetic force on the electron is given by

#### Answer:

Given:
For the outer loop,
Magnitude of current = I
Radius of the loop = R
Thus, the magnetic field at the centre due to the larger loop is given by
$B=\frac{{\mu }_{0}I}{2R}$
Let A be the area of the smaller loop and let current i pass through it.
Now,
Angle between the area vector of the smaller loop and the magnetic field due to the larger loop = 90°
Thus, the required torque is given by

= iABsin 90°
$=i\mathrm{\pi }{r}^{2}\frac{{\mathrm{\mu }}_{0}I}{2R}\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{\mu }}_{0}\mathrm{\pi }{r}^{2}Ii}{2R}$

#### Answer:

Given:
For the outer loop,
Magnitude of current = I
Radius of the loop = R
Thus, the magnetic field at the centre due to the larger loop is given by

Let A be the area of the smaller loop and let current i pass through it.
Now,
Angle between the area vector of the smaller loop and the magnetic field due to the larger loop = 30°
Thus, the torque on the smaller loop is given by

= iABsin 30°
$=i\mathrm{\pi }{r}^{2}\frac{{\mathrm{\mu }}_{0}I}{4R}\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{\mu }}_{0}\mathrm{\pi }{r}^{2}Ii}{4R}$
If the smaller loop is held fixed in its position, then
Torque due to the magnetic field = Torque due to the external force at its periphery

This is the minimum magnitude of force to balance the given condition.

#### Answer:

Given:
Magnitude of current, I = 5 A
Radius of the semi-circular wire, r = 10 cm
∴ Required magnetic field at the centre of curvature

#### Answer:

Given:
Magnitude of current, I = 6 A
Radius of the semi-circular wire, r = 10 cm
Angle subtended at the centre, θ = 120° = $\frac{2\mathrm{\pi }}{3}$
∴ Required magnetic field at the centre of curvature

#### Answer:

Given:
Magnitude of current = i
Radius of the loop = r
Magnetic field due to the loop at its centre,
Let a straight wire carrying 4i current be placed at a distance x from the centre such that the magnetic fields of the loop and the wire are of equal magnitude but in opposite direction at O.

Magnetic field due to the wire at the centre of the loop,
According to the question,

This means that the wire is placed $\frac{4r}{\mathrm{\pi }}$ from the centre of the loop (as shown in the figure).

#### Answer:

Number of turns, n = 200
Radius of the coil, r = 10 cm
Current in the coil, i = 2A
(a) Let the magnetic field at the centre of the coil is B.
As the relation for magnetic field at the centre of a circular coil is given by

(b) As magnetic field at any point P (say) on the axis of the circular coil is given by

Where x is the distance of the point from the centre of the coil.
As per the question

Magnetic field will drop to half of its value at the centre if the distance of that point from the centre of the coil along the axis of coil is equal to 7.66 cm.

#### Answer:

Given:
Magnitude of current, I = 5.0 A
Radius of the loop, r = 4.0 cm

(a) The magnetic field intensity B on point O at a distance x on the axial line is given by

(b) The magnetic field intensity B on point O' at a distance x on the axial line is given by

#### Answer:

Given:
Magnitude of charges, q =  3.14 × 10−6 C
Radius of the ring,
Angular velocity of the ring,
Time for 1 revolution = $\frac{2\mathrm{\pi }}{60}$

In the figure, E1 and E2 denotes the electric field at a point on the axis at a distance of 5.00 cm from the centre due to small element 1 and 2 of the ring respectively.
E is the resultant electric field due to the entire ring at a point on the axis at a distance of 5.00 cm from the centre.
The electric field at a point on the axis at a distance x from the centre is given by

The magnetic field at a point on the axis at a distance x from the centre is given by

#### Answer:

(a) The magnetic field inside any conducting tube is always zero.
∴ Magnetic field inside the tube at a distance r/2 from the surface = 0

(b) Let the point outside the tube with distance $\frac{r}{2}$ be P.

∴ Net distance from centre, r' = $r+\frac{r}{2}=\frac{3r}{2}$
Consider an Amperian loop, as shown in the figure.
Length of the loop, l = $2\mathrm{\pi }×\frac{3}{2}r=3\mathrm{\pi }r$
Current enclosed in the loop = i
On applying Ampere's law, we get
$\int B.dl={\mu }_{0}i\phantom{\rule{0ex}{0ex}}⇒B×3\mathrm{\pi }r={\mu }_{0}i\phantom{\rule{0ex}{0ex}}⇒B=\frac{{\mu }_{0}i}{3\mathrm{\pi }r}\phantom{\rule{0ex}{0ex}}$

#### Answer:

a) The magnetic field inside any conducting tube is always zero.
∴ Magnetic field just inside the tube is zero.

(b) Let the point outside the tube with distance b be P.
Consider an Amperian loop, as shown in the figure.
Length of the loop, l = $2\mathrm{\pi }×b=2\mathrm{\pi }b$
Current enclosed in the loop = i
On applying Ampere's law, we get
$\int B.dl={\mu }_{0}i\phantom{\rule{0ex}{0ex}}⇒B×2\mathrm{\pi }b={\mu }_{0}i\phantom{\rule{0ex}{0ex}}⇒B=\frac{{\mu }_{0}i}{2\mathrm{\pi }b}\phantom{\rule{0ex}{0ex}}$

#### Answer:

Given:
Magnitude of current = i
Radius of the wire = b

For a point at a distance a from the axis,
Current enclosed, $i\text{'}=\frac{i}{\mathrm{\pi }{b}^{2}}×\mathrm{\pi }{a}^{2}$
By Ampere's circuital law,

For the given conditions,

#### Answer:

Given:
Magnitude of current, i = 5 A
Radius of the wire, b
For a point at a distance a from the axis,
Current enclosed,
By Ampere's circuital law,

For the given conditions,

Again, using the circuital law, we get

(b) On putting in (1), we get
B = 10 $\mathrm{\mu T}$

(c)Using the circuital law, we get

#### Answer:

Half of the loop PQRS is in the region of magnetic field and half in the area of zero magnetic field.

Let us consider a current carrying circular wire, due to which there is uniform magnetic field in the region.

Take a point A inside the loop PQRS in the region where B = 0

According to Ampere's circuital law,
$\int \mathrm{B}.dl={\mu }_{0}i$

If there is current enclosed by the loop PQRS, then magnetic field B cannot be 0.

Whereas, we have taken the magnetic field at point A to be zero.

Thus, such a field is not possible.

#### Answer:

At point P,
Current, i = 0
∴ Magnetic field B = 0
At point Q,
Applying Ampere's law, we get

At point R,
Current, i = 0
∴ B = 0

#### Answer:

Given:
Charge = q
Mass = m
Radius = r
We know that the radius described by a charged particle in a magnetic field is given by
$r=\frac{mv}{q\mathrm{B}}\phantom{\rule{0ex}{0ex}}$

$⇒v=\frac{\mathrm{B}qr}{m}=\frac{{\mu }_{0}kqr}{m}$

#### Answer:

Given:
Magnitude of current, i = 5 A
Magnetic field intensity, B = 3.14 × 10−2 T
We know that the magnetic field inside a long solenoid having n turns per unit length is given by

#### Answer:

Given:
Radius of the wire, r = 0.5 mm
Width of each turn, = diameter of the wire, 2r = 1 mm = 1 × 10−3 m
∴ Total number of turns in 1 m solenoid

#### Answer:

Given:
Resistance per unit length of the wire, $\frac{R}{l}$ = 0.01 Ω/m
Radius of the wire, r = 1.0 cm = 0.01 m
Total no. of turns, N = 400
Magnetic field intensity, B = 1.0 × 10−2 T
Now,
Let E be the emf of the battery and R0 be the total resistance of the wire.
$\therefore i=\frac{E}{{R}_{0}}=\frac{E}{0.01×2\mathrm{\pi }r×400}\phantom{\rule{0ex}{0ex}}=\frac{E}{0.01×2×\mathrm{\pi }×0.01×400}$

The magnetic field near the centre of the solenoid is given by

#### Answer:

(a) Given:
Current in the loop or circular current = indx
Radius of the loop having circular current  = r
Distance of the centre of the solenoid from the circular current = $\frac{l}{2}-x$
Magnetic field at the centre due to the circular loop,
$B=\frac{{\mu }_{0}}{2}\frac{i{r}^{2}}{\left({x}^{2}+{r}^{2}{\right)}^{3/2}}$
$\mathrm{B}=\int d\mathrm{B}\phantom{\rule{0ex}{0ex}}=\underset{0}{\overset{1}{\int }}\frac{{\mathrm{\mu }}_{0}{a}^{2}nidx}{4\pi {\left[{a}^{2}+\left(l-2x{\right)}^{2}\right]}^{3/2}}\phantom{\rule{0ex}{0ex}}=\underset{0}{\overset{1}{\int }}\frac{{\mathrm{\mu }}_{0}ni{a}^{2}dx}{4\pi {a}^{3}{\left[1+{\left(\frac{l-2x}{a}\right)}^{2}\right]}^{3/2}}=\frac{{\mu }_{0}ni}{4\pi a}\underset{0}{\overset{1}{\int }}\frac{dx}{{\left[1+{\left(\frac{l-2x}{a}\right)}^{2}\right]}^{3/2}}=\frac{{\mu }_{0}ni}{4\pi a}.\frac{4\pi a}{\sqrt{1+\left(\frac{2a}{l}\right)}}=\frac{{\mu }_{0}ni}{\sqrt{1+{\left(\frac{2a}{l}\right)}^{2}}}$
(b) When a > > l,
$\mathrm{B}=\frac{{\mu }_{0}ni}{2a}$

#### Answer:

Given:
Magnitude of current in the solenoid, i = 2 A
Frequency of the electron,
Mass of the electron,
Charge of the electron,
We know that the magnetic field inside a solenoid is given by
B = µ0ni
If a particle executes uniform circular motion inside a magnetic field, the frequency of the particle is given by

#### Answer:

Given:
Magnitude of current in the solenoid = i
Number of turns per unit length = n
When a particle is projected perpendicular to the magnetic field, it describes a circular path.
And for the particle (projected from a point on the axis in a direction perpendicular to the axis) to not strike the solenoid, the maximum radius of that circular path should be r/2.

∴ Radius of the circle = $\frac{r}{2}$

We know,
Centripetal force = Magnetic force
$\frac{m{V}^{2}}{r}=qVB\phantom{\rule{0ex}{0ex}}⇒V=\frac{qBr}{m}=\frac{q{\mu }_{0}nir}{2m}$

#### Answer:

Given:
Number of turns per unit length of the solenoid = n
(a) Since the net magnetic field near the centre of the solenoid is 0,

For the solenoid,
${B}_{\mathrm{solenoid}}={\mathrm{\mu }}_{0}ni$    ...(2)

From (1) and (2), we get

(b) On putting the value of n in (2), we get

Now, are perpendicular to each other.
Thus, the net magnetic field near the centre of the solenoid is given by

#### Answer:

Given:
Capacitance, C = 100 microfarad
Voltage, V = 20 V
Charge stored in the capacitor, Q = CV

It is given that the potential difference across the capacitor drops to 90% of its maximum value.
Thus,

No. of turns per metre, n = 4000
Thus, the average magnetic field at the centre of the solenoid is given by

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