Question

Two long straight parallel wires carry currents of 8 A and 10 A in opposite directions. If the distance of separation between the wires is 9 cm, then the net magnetic field at a point between the two wires, which is at a perpendicular distance of 4 cm from the wire carrying 8 A current is

• A. Zero
• B. \(4 \times 10^{-5}\) T
• C. \(8 \times 10^{-5}\) T
• D. \(12 \times 10^{-5}\) T

Hint:

Use the right-hand grip rule to determine the direction of magnetic fields. For two parallel wires: - Currents in the same direction: Fields oppose each other in between the wires. - Currents in opposite directions: Fields add up in between the wires.

Answer 1

The Correct Option is C

The magnetic field (B) at a distance r from a long straight wire carrying current I is given by Ampere's law: \( B = \frac{\mu_0 I}{2\pi r} \).
The value of the constant \( \frac{\mu_0}{2\pi} \) is \( 2 \times 10^{-7} \) T·m/A.
Let's call the wire with 8 A current wire 1, and the wire with 10 A current wire 2.
\( I_1 = 8 \) A, \( I_2 = 10 \) A. The currents are in opposite directions.
The point P is between the wires. The distance from wire 1 is \( r_1 = 4 \text{ cm} = 0.04 \text{ m} \).
The total separation is 9 cm, so the distance from wire 2 is \( r_2 = 9 - 4 = 5 \text{ cm} = 0.05 \text{ m} \).
At a point between two wires carrying opposite currents, the magnetic fields from both wires point in the same direction (by the right-hand grip rule). Therefore, the net magnetic field is the sum of the individual fields.
Magnetic field from wire 1: \( B_1 = \frac{\mu_0 I_1}{2\pi r_1} = (2 \times 10^{-7}) \frac{8}{0.04} = (2 \times 10^{-7}) \times 200 = 400 \times 10^{-7} = 4 \times 10^{-5} \) T.
Magnetic field from wire 2: \( B_2 = \frac{\mu_0 I_2}{2\pi r_2} = (2 \times 10^{-7}) \frac{10}{0.05} = (2 \times 10^{-7}) \times 200 = 400 \times 10^{-7} = 4 \times 10^{-5} \) T.
The net magnetic field is the sum of \(B_1\) and \(B_2\).
\( B_{net} = B_1 + B_2 = 4 \times 10^{-5} + 4 \times 10^{-5} = 8 \times 10^{-5} \) T.