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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) 3BThe 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»

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


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