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

To solve this problem, we need to calculate the magnetic field produced at a point between two current-carrying wires. The wires carry currents in opposite directions, and the point is located closer to the wire carrying 8 A.

Let's denote:

• \(I_1 = 8 \, \text{A}\) (current in the first wire)
• \(I_2 = 10 \, \text{A}\) (current in the second wire)
• \(d = 9 \, \text{cm} = 0.09 \, \text{m}\) (distance between the wires)
• \(r_1 = 4 \, \text{cm} = 0.04 \, \text{m}\) (distance from point to first wire)
• \(r_2 = d - r_1 = 0.09 - 0.04 = 0.05 \, \text{m}\) (distance from point to second wire)

The formula for the magnetic field at a distance \(r\) from a long straight wire carrying current \(I\) is given by:

\(B = \frac{\mu_0 I}{2 \pi r}\)

Where \(\mu_0 = 4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A}\) is the permeability of free space.

Step 1: Calculate the magnetic field \(B_1\) due to the 8 A wire at the point.

\(B_1 = \frac{4\pi \times 10^{-7} \times 8}{2 \pi \times 0.04} = \frac{32 \times 10^{-7}}{0.08} = 4 \times 10^{-5} \, \text{T}\)

Step 2: Calculate the magnetic field \(B_2\) due to the 10 A wire at the point.

\(B_2 = \frac{4\pi \times 10^{-7} \times 10}{2 \pi \times 0.05} = \frac{40 \times 10^{-7}}{0.10} = 4 \times 10^{-5} \, \text{T}\)

Step 3: Determine the net magnetic field.

Since the currents are in opposite directions, the magnetic fields at the point will subtract.

\(B_{\text{net}} = B_2 - B_1 = 4 \times 10^{-5} - 4 \times 10^{-5} = 0 \, \text{T}\)

However, reviewing set calculations indicates an error in subtraction because the initial explanation assumed subtraction. Therefore, recalculate by logical arrangement of distances:

• The field due to 10A wire is from further distance position away, and directional diference by \(\angle\) would solidify:
• Thus, knowing a logical oversight gets variation retried:

Correct resultant:

\(B_{\text{net}} = B_2 + B_1 = 4 \times 10^{-5} + 4 \times 10^{-5} = 8 \times 10^{-5} \, \text{T}\)

Therefore, the correct answer is the net magnetic field between the two wires is \(8 \times 10^{-5} \, \text{T}\).