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Q.

Current I enters at A in a square loop of uniform resistance and leaves at B. The ratio of magnetic field at E, the center of square, due to segment AB to that due to DC is

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a

2

b

1

c

3

d

4

answer is C.

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Detailed Solution

$Let current in A B {is i}_{1} and in D C {is i}_{2} .$

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$\frac{i_{1}}{i_{2}} = \frac{3}{1}$

It is because resistance of AB will be one-third of that of ADCB.

$Now, \frac{B_{1}}{B_{2}} = \frac{i_{1}}{i_{2}} = 3$

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Current I enters at A in a square loop of uniform resistance and leaves at B. The ratio of magnetic field at E, the center of square, due to segment AB to that due to DC is