Question

Two identical long parallel wires carry current '\(I_{1}\)' and '\(I_{2}\)' such that \(I_{1}>I_{2}\). When the currents are in the same direction, the magnetic field at a point midway between the wires is \(6\times 10^{-6}\ \text{T}\). If the direction of '\(I_{2}\)' is reversed, the field becomes \(3\times 10^{-5}\ \text{T}\). The ratio \(\left(\frac{I_{1}}{I_{2}}\right)\) is

• A. \(\frac{1}{2}\)
• B. \(\frac{2}{3}\)
• C. \(\frac{3}{4}\)
• D. \(\frac{3}{2}\)

Hint:

At the midpoint, same-direction currents give opposing fields, and opposite currents give fields that add.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
At the midpoint each long wire gives a field \(B=\dfrac{\mu_{0}I}{2\pi d}\). Let \(k=\dfrac{\mu_{0}}{2\pi d}\) with \(d\) the distance from each wire.

Step 2: Both conditions
Same direction: the fields oppose, \(k(I_{1}-I_{2})=6\times10^{-6}\ \text{T}\).
When \(I_{2}\) is reversed: the fields add, \(k(I_{1}+I_{2})=3\times10^{-5}\ \text{T}\).

Step 3: Calculation
Dividing: \(\dfrac{I_{1}+I_{2}}{I_{1}-I_{2}}=5\), so \(I_{1}+I_{2}=5I_{1}-5I_{2}\), giving \(6I_{2}=4I_{1}\) and \(\dfrac{I_{1}}{I_{2}}=\dfrac{3}{2}\).

Final Answer:
The ratio I1/I2 is 3/2. \[ \boxed{\text{(D) }\dfrac{3}{2}} \]