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

Options

A. `1:1`
B. `1:2`
C. `1:3`
D. `1:4`

Answer

C

Explanation

(c) Let current in AB is `I_1` and in DC,`I_2`. Then

`(I_1)/(I_2)=3/1`. It is because resistance of AB will be one-third of
that of ADCB.
Now, `(B_1)/(B_2)=(I_1)/(I_2)=3`