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Question

# There is a square loop of side l=10 cm. The resistance of loop is 0.5 Ω/cm. The loop is moved with constant velocity 2 ^i m/s in two uniform magnetic field as shown in the figure. The magnitude of magentic field is 10 T in both regions. Then

A
The potential difference across the points CD is zero
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B
The potential difference across the points AC is 2 V
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C
Current through the loop is 20 A
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D
Current through the loop is 10 A
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Solution

## The correct option is C Current through the loop is 20 AAs length is parallel to velocity for sides CD and AB, emf generated in these sides is 0 For Side BC, emf generated is Blv=10(0.1)2=2 V Potential difference across AC is 2+0=2 V For Side AD, emf generated is Blv=10(0.1)2=2 V Net potential in the loop is V=4 V Total resistance of the loop is 0.5(4×0.1)=0.2 Ω Current through the loop is VR=20 A

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