Question

A long, straight wire carries a current of 10 A. The magnitude of the magnetic field at a distance of 5 cm from the wire is:

• A. \(4 \times 10^{-5}\) T
• B. \(8 \times 10^{-5}\) T
• C. \(12 \times 10^{-5}\) T
• D. \(16 \times 10^{-5}\) T

Hint:

The expression \(\frac{\mu_0}{2\pi}\) can be simplified to \(2 \times 10^{-7}\) T\(\cdot\)m/A, which makes calculations for the magnetic field of a straight wire quicker.

Answer 1

The Correct Option is A

Step 1: Recall the formula for the magnetic field of a long, straight wire. Ampere's Law gives the magnetic field \(B\) at a perpendicular distance \(r\) from a long, straight wire carrying a current \(I\): \[ B = \frac{\mu_0 I}{2\pi r} \] where \(\mu_0\) is the permeability of free space, \(\mu_0 = 4\pi \times 10^{-7}\) T\(\cdot\)m/A.
Step 2: Identify the given values and convert them to SI units. - Current \(I = 10\) A. - Distance \(r = 5 \, \text{cm} = 0.05 \, \text{m}\).
Step 3: Substitute the values into the formula and calculate \(B\). \[ B = \frac{(4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A}) \times (10 \, \text{A})}{2\pi \times (0.05 \, \text{m})} \] \[ B = \frac{2 \times 10^{-7} \times 10}{0.05} \, \text{T} = \frac{2 \times 10^{-6}}{5 \times 10^{-2}} \, \text{T} \] \[ B = \frac{2}{5} \times 10^{-4} \, \text{T} = 0.4 \times 10^{-4} \, \text{T} = 4 \times 10^{-5} \, \text{T} \]