Question

A current of \(30\,\text{A}\) each flows in opposite directions in two conducting wires, placed parallel to each other at a distance of \(8\,\text{cm}\). The magnetic field at the mid point between the two wires is _____ \(\mu\text{T}\). \(\left(\frac{\mu_0}{4\pi}=10^{-7}\,\text{}^2\right)\)

• A. \(30\)
• B. \(300\)
• C. \(150\)
• D. \(0\)

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The objective is to find the total magnetic field at the center of two parallel wires carrying current in opposite directions.
Step 2: Key Formula or Approach:
The magnetic field produced by a long straight wire at a distance \(r\) is:
\[ B = \frac{\mu_0 I}{2\pi r} \]
By the right-hand thumb rule, if currents are in opposite directions, their magnetic fields add up at the midpoint between them.
Step 3: Detailed Explanation:
Current \(I = 30 \text{ A}\). Separation \(d = 8 \text{ cm}\).
Distance to midpoint \(r = \frac{d}{2} = 4 \text{ cm} = 0.04 \text{ m}\).
Since fields add:
\[ B_{net} = B_1 + B_2 = 2 \times \frac{\mu_0 I}{2\pi r} = \frac{\mu_0 I}{\pi r} \]
Substitute values:
\[ B_{net} = \frac{(4\pi \times 10^{-7}) \times 30}{\pi \times 0.04} \]
\[ B_{net} = \frac{4 \times 10^{-7} \times 30}{0.04} = \frac{120 \times 10^{-7}}{4 \times 10^{-2}} \]
\[ B_{net} = 30 \times 10^{-5} \text{ T} = 300 \times 10^{-6} \text{ T} = 300 \text{ }\mu\text{T} \]
Step 4: Final Answer:
The magnetic field at the midpoint is \(300 \text{ }\mu\text{T}\).