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: Write the expression for magnetic field on the equatorial line.
For a short bar magnet, the magnetic field at a point on the normal bisector (equatorial line) is \[ B_e=\frac{\mu_0}{4\pi}\frac{M}{r^3} \] Given, \[ B_e=6.4\times 10^{-5}\,\text{T} \] at \[ r_e=20\,\text{cm}=0.2\,\text{m} \]

Step 2: Write the expression for magnetic field on the axial line.
For a short bar magnet, the magnetic field at a point on the axis is \[ B_a=\frac{\mu_0}{4\pi}\frac{2M}{r^3} \] We have to find the field at \[ r_a=40\,\text{cm}=0.4\,\text{m} \]

Step 3: Take the ratio of axial and equatorial fields.
Dividing the two expressions, \[ \frac{B_a}{B_e} = \frac{\dfrac{\mu_0}{4\pi}\dfrac{2M}{r_a^3}} {\dfrac{\mu_0}{4\pi}\dfrac{M}{r_e^3}} \] \[ \frac{B_a}{B_e} = 2\left(\frac{r_e}{r_a}\right)^3 \] Substituting the values, \[ \frac{B_a}{B_e} = 2\left(\frac{0.2}{0.4}\right)^3 \] \[ \frac{B_a}{B_e} = 2\left(\frac{1}{2}\right)^3 \] \[ \frac{B_a}{B_e} = 2\times \frac{1}{8} \] \[ \frac{B_a}{B_e} = \frac{1}{4} \] Therefore, \[ B_a=\frac{B_e}{4} \] \[ B_a=\frac{6.4\times 10^{-5}}{4} \] \[ B_a=1.6\times 10^{-5}\,\text{T} \]

Step 4: Final conclusion.
Therefore, the magnetic field at \(40\,\text{cm}\) on the axis is \[ \boxed{1.6\times 10^{-5}\,\text{T}} \]