Question:
The magnitude of the magnetic field produced by a short bar magnet at a distance of 20 cm from the centre of the magnet on the normal bisector of the magnet is found to be \(5 \times 10^{-6} \, T\). The magnetic moment of the bar magnet is:
The magnitude of the magnetic field produced by a short bar magnet at a distance of 20 cm from the centre of the magnet on the normal bisector of the magnet is found to be \(5 \times 10^{-6} \, T\). The magnetic moment of the bar magnet is:
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On normal bisector, field \(B = (\mu_0/4\pi)(M/r^3)\). Solve for magnetic moment by rearranging formula.
Updated On: Jul 18, 2026
- 0.1 J T\(^{-1}\)
- 0.4 J T\(^{-1}\)
- 0.6 J T\(^{-1}\)
- 1.2 J T\(^{-1}\)
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The Correct Option is B
Solution and Explanation
Step 1: Formula for field on normal bisector.
Magnetic field at a point on the perpendicular bisector (normal line) of a bar magnet of magnetic moment \(M\) is: \[ B = \frac{\mu_0}{4\pi} \frac{M}{r^3} \]
Step 2: Identify given values.
\[ B = 5 \times 10^{-6} \, T, \quad r = 0.2 \, m, \quad \frac{\mu_0}{4\pi} = 10^{-7} \, SI \]
Step 3: Solve for magnetic moment \(M\).
\[ M = \frac{B r^3}{\mu_0/4\pi} = \frac{5 \times 10^{-6} \cdot (0.2)^3}{10^{-7}} \]
Step 4: Compute \(r^3\).
\[ 0.2^3 = 0.008 \] \[ 5 \times 10^{-6} \cdot 0.008 = 4 \times 10^{-8} \]
Step 5: Divide by \(10^{-7}\).
\[ M = \frac{4 \times 10^{-8}}{10^{-7}} = 0.4 \, J T^{-1} \]
Step 6: Final conclusion.
\[ \boxed{0.4 \, J T^{-1}} \]
Magnetic field at a point on the perpendicular bisector (normal line) of a bar magnet of magnetic moment \(M\) is: \[ B = \frac{\mu_0}{4\pi} \frac{M}{r^3} \]
Step 2: Identify given values.
\[ B = 5 \times 10^{-6} \, T, \quad r = 0.2 \, m, \quad \frac{\mu_0}{4\pi} = 10^{-7} \, SI \]
Step 3: Solve for magnetic moment \(M\).
\[ M = \frac{B r^3}{\mu_0/4\pi} = \frac{5 \times 10^{-6} \cdot (0.2)^3}{10^{-7}} \]
Step 4: Compute \(r^3\).
\[ 0.2^3 = 0.008 \] \[ 5 \times 10^{-6} \cdot 0.008 = 4 \times 10^{-8} \]
Step 5: Divide by \(10^{-7}\).
\[ M = \frac{4 \times 10^{-8}}{10^{-7}} = 0.4 \, J T^{-1} \]
Step 6: Final conclusion.
\[ \boxed{0.4 \, J T^{-1}} \]
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