Question

A short bar magnet produces a magnetic field of \(6.4\times 10^{-5}\,\text{T}\) at a distance of \(20\,\text{cm}\) from the center of the magnet on the normal bisector of the magnet. The magnetic field produced by this magnet at a distance of \(40\,\text{cm}\) from the center of the magnet on the axis is:

• A. \(4.8\times 10^{-5}\,\text{T}\)
• B. \(3.2\times 10^{-5}\,\text{T}\)
• C. \(1.6\times 10^{-5}\,\text{T}\)
• D. \(6.4\times 10^{-5}\,\text{T}\)

Hint:

For a short bar magnet, \[ B_{\text{axial}}=\frac{\mu_0}{4\pi}\frac{2M}{r^3} \] and \[ B_{\text{equatorial}}=\frac{\mu_0}{4\pi}\frac{M}{r^3} \] Thus, at the same distance, \[ B_{\text{axial}}=2B_{\text{equatorial}}. \]

Answer 1

The Correct Option is C

Step 1: Extract the magnetic moment from the equatorial field.
\( B_e = \frac{\mu_0}{4\pi}\frac{M}{r_e^3} \). So \( \frac{\mu_0 M}{4\pi} = B_e r_e^3 = 6.4\times10^{-5}\times(0.2)^3 = 5.12\times10^{-7}\,\text{T m}^3 \).

Step 2: Compute axial field at r = 0.4 m using \( B_{axis} = \frac{\mu_0}{4\pi}\frac{2M}{r^3} \).
\[ B_{axis} = \frac{2\times5.12\times10^{-7}}{(0.4)^3} = \frac{10.24\times10^{-7}}{0.064} = 1.6\times10^{-5}\,\text{T} \] \[ \boxed{1.6\times10^{-5}\,\text{T}} \]