Question

Two identical long parallel wires carry currents ' \(\mathrm{I}_1\) ' and ' \(\mathrm{I}_2\) ' such that \(\mathrm{I}_1>\mathrm{I}_2\). When the currents are in the same direction, the magnetic field at a point midway between the wires is \(8 \times 10^{-6} \mathrm{~T}\). If the direction of \(\mathrm{I}_2\) is reversed, the field becomes \(3.2 \times 10^{-5} \mathrm{~T}\). The ratio of \(\mathrm{I}_2\) to \(\mathrm{I}_1\) is

• A. \(1:4\)
• B. \(2:5\)
• C. \(3:5\)
• D. \(3:4\)

Hint:

At the midpoint the fields of the two wires subtract for same-direction currents and add for opposite currents.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
At the midpoint each wire gives \(B=\dfrac{\mu_0 I}{2\pi d}\). With currents in the same direction the fields oppose; with opposite currents they add.

Step 2: Two equations
Let \(k=\dfrac{\mu_0}{2\pi d}\). Then \(k(I_1-I_2)=8\times10^{-6}\) and \(k(I_1+I_2)=3.2\times10^{-5}\).

Step 3: Calculation
\(\dfrac{I_1+I_2}{I_1-I_2}=4\Rightarrow I_1+I_2=4I_1-4I_2\Rightarrow5I_2=3I_1\)
\(\dfrac{I_2}{I_1}=\dfrac{3}{5}\)

Final Answer:
The ratio of I2 to I1 is 3 : 5. \[ \boxed{\text{(C) }3:5} \]