Question

Two parallel long wires 'A' and 'B' carry currents \( i_1 \) and \( i_2 \) (\( i_1<i_2 \)). When \( i_1 \) and \( i_2 \) are in the same direction, the magnetic field at a point midway between the wires is 10 µT. If \( i_2 \) is reversed, then the field becomes 30 µT. The ratio of \( \frac{i_1}{i_2} \) is

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

When dealing with magnetic fields due to current-carrying wires, use the formula \( B = \frac{\mu_0 I}{2\pi d} \) and apply superposition for adding or subtracting fields from multiple currents.

Answer 1

The Correct Option is B

Step 1: Magnetic field due to current-carrying wire.
The magnetic field at a point due to a current \( I \) in a long straight wire is given by: \[ B = \frac{\mu_0 I}{2\pi d} \] where \( \mu_0 = 4\pi \times 10^{-7} \, \text{T·m/A} \) is the permeability of free space and \( d \) is the distance from the wire.
Step 2: Using the given data.
When both currents are in the same direction, the magnetic fields due to both wires add up, and at the point midway between them, the total field is: \[ B_{\text{same}} = \frac{\mu_0 i_1}{2\pi d} + \frac{\mu_0 i_2}{2\pi d} \] Given \( B_{\text{same}} = 10 \, \mu T \). When \( i_2 \) is reversed, the magnetic fields due to the two currents subtract: \[ B_{\text{opposite}} = \frac{\mu_0 i_2}{2\pi d} - \frac{\mu_0 i_1}{2\pi d} \] Given \( B_{\text{opposite}} = 30 \, \mu T \). Step 3: Solving for the ratio.
From these two equations, we can solve for the ratio \( \frac{i_1}{i_2} \) and find it to be 2.
Step 4: Conclusion.
Thus, the ratio \( \frac{i_1}{i_2} \) is 2, corresponding to option (2).