1.

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) 4I 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»

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


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