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NCERT Class XII Chapter
Electromagnetic Induction
Questions With Solutions

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Question : 12
Total: 17
A square loop of side 12 cm with its sides parallel to X and Y axes is moved with a velocity of 8 cm s–1 in the positive x-direction in an environment containing a magnetic field in the positive z-direction. The field is neither uniform in space nor constant in time. It has a gradient of 10–3Tcm–1 along the negative x-direction (that is it increases by 10–3Tcm–1 as one moves in the negative x-direction) and it is decreasing in time at the rate of 10–3 T s–1. Determine the direction and magnitude of the induced current in the loop if its resistance is 4.50 mΩ
Solution:  

Each side of square loop is 12 cm and magnetic field is decreasing along x direction.
dB
dx
 = - 10−3Tcm−1 = - 0.1 T m−1
Also the magnetic field is decreasing with time at constant rate
dB
dt
 = - 10−3Ts−1
Induced emf and rate of change of magnetic flux due to only time variation
εt = -
dϕ
dt
 = -
dBA
dt
 = - A
dB
dt
 or εt = - 0.12 × 0.12 [−10−3] = 144 × 10−7 V
Induced emf and rate of change of magnetic flux due to change in position.
εx = -
dBA
dt
 = - A
dB
dx
 ×
dx
dt
εx = - Av
dB
dx
 = -0.12 × 0.12 × 0.08 × (0.1) = 1152 × 10−7 V
Both the induced emf have same sign and thus adds to provide net Induced emf in the loop
εnet = εi+εx = 1296 × 10−7 V
Induced current
I =
εnet
R
 =
1296×10−7
4.5×10−3
 = 2.88 × 10−2 A
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