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

Concept: Magnetic field due to a long straight current carrying wire: \[ B=\frac{\mu_0 I}{2\pi r} \] If currents flow in opposite directions, magnetic fields at the midpoint add. Step 1: {Determine distance from each wire.} Distance between wires: \[ d=8\,\text{cm} \] Distance from midpoint to each wire: \[ r=4\,\text{cm}=0.04\,\text{m} \] Step 2: {Magnetic field due to one wire.} \[ B=\frac{\mu_0 I}{2\pi r} \] Since \[ \frac{\mu_0}{4\pi}=10^{-7} \] \[ \frac{\mu_0}{2\pi}=2\times10^{-7} \] Thus \[ B=\frac{2\times10^{-7}\times30}{0.04} \] \[ =1.5\times10^{-4}\,\text{T} \] Step 3: {Add fields from both wires.} Since fields are in same direction: \[ B_{total}=2\times1.5\times10^{-4} \] \[ =3\times10^{-4}\,\text{T} \] Convert to \(\mu\text{T}\): \[ 3\times10^{-4}\times10^6 \] \[ =300\,\mu\text{T} \]