Question

There are 2 long parallel conductors AB and CD. AB carries 4A current and CD carries 3A current. The magnetic field at the midpoint of these 2 conductors is B. If 4A current is switched off, then what is the magnetic field at the midpoint now?

(a) \(\frac {B}{3}\)

(b) \(\frac {2B}{3}\)

(c) B

(d) 3B

The question was asked during a job interview.

My question is from Ampere and Forces between Two Parallel Currents in chapter Moving Charges and Magnetism of Physics – Class 12

Answer 1

The correct option is (d) 3B

For explanation: Firstly, the magnetic field at the midpoint due to current in AB:

BAB=μo×\(\frac {2 \times 4}{4\pi d}=\frac {8\mu_o}{4\pi d}\) ………………..1

Magnetic field at midpoint due to current in CD:

BCD=μo×\(\frac {2 \times 3}{4\pi d}=\frac {6\mu_o}{4\pi d}\) …………………..2

The net magnetic field = BAB – BCD ➔ 1 – 2

B = \(\frac {2 \mu_o}{4\pi d}\) ……………………….3

Now, when 4A is switched off, then the magnetic field at the midpoint will be due to the current in CD, i.e. due to 3A current

B^‘ = μo×\(\frac {2 \times 3}{4\pi d}=\frac {6\mu_o}{4\pi d}\) ……………..4

Comparing 3 and 4:

Therefore, B^‘ = 3B