The magnetic field through a single loop of wire, 12cm in radius and 8.5 ohm resistance changes with time as shown. The magnetic field is perpendicular to the plane of the loop plot induced current as a function of time.

Dear student,

The induced emf of loop is given by  ε=-dφmdt ,

Flux is given by φm =B(t)×Area

Lets find B(t) for all the time interval.
For t1 =0s to t2 =2s
Its a line .
B=mt+b,b=0 as it starts from origin 
1=2m
m= 0.5 
B=0.5T.
Magnetic flux is given by

 φm=B(t) × areaφm =0.5T×π×r2φm=0.5T×3.14×0.12 m×0.12 m=0.023emf=-dt=-0.0232=-0.0115 V=-1.15×10-2V

From t=2 s to t=4 s,The magnetic field  is constant.Hence dBdt=0.emf =0 V
From t=4s to t= 6 S,Its a straight line.

B(t)= m t+b, 
0=6m+b0.5=4m+b2m=-0.5 ,m=-0.25,b=3B(t)=-0.25T+3
The magnetic flux is

 φm=B(t)×areaφm=(-0.25T+3)×3.14×0.12m×0.12 mφm =2.75×0.045=0.123emf =-dφdt=-0.1232=-0.061V=-6×10-2 V
The current is given by ohms law I =vR
For time t =0s to t=2s 
I =1.15×10-28.5 =-1.35×10-3A
For time t=2s to t=4s ,I=0
For time t=4s to t=6s,
I=-6×10-28.5=-7.05×10-3A


Regards.



 

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