Question

Two parallel, long wires carry currents I1 and I2 where I1 > I2. When the currents are in the same direction, the magnetic field at a point midway between the wires is 50 μT. If the direction of I2 is revered, the field becomes 100 μT. What is the value of \(\frac {I_1}{I_2}\)?

(a) 1

(b) 2

(c) 3

(d) 4

I have been asked this question in an interview for internship.

I want to ask this question from Ampere and Forces between Two Parallel Currents in division Moving Charges and Magnetism of Physics – Class 12

Answer 1

Correct choice is (c) 3

To explain: At the midpoint ➔ CASE 1:

B1=\(\frac {\mu_o I_1}{2\pi d} – \frac {\mu_o I_2}{2\pi d}\)

B1=\(\frac {\mu_o (I_1 – I_2)}{2\pi d}\)=50 μT

At the midpoint ➔ CASE 2:

B2=\(\frac {\mu_o I_1}{2\pi d} + \frac {\mu_o I_2}{2\pi d}\)

B1=\(\frac {\mu_o (I_1 + I_2)}{2\pi d}\)=100 μT

\(\frac {B1}{B2}=\frac {I_1-I_2}{I_1+I_2}=\frac {50 \mu T}{100 \mu T}\)

\(\frac {I_1-I_2}{I_1+I_2}=\frac {1}{2}\)

Using componendo and Dividendo rule:

\(\frac {(I_1+I_2) \, + \, (I_1-I_2)}{(I_1-I_2) \, – \, (I_1+I_2)}=\frac {1+2}{1-2}\)

\(\frac {I_1}{-I_2}=\frac {3}{-1}\)

Therefore, \(\frac {I_1}{I_2}\)=3