Question

A cylindrical magnetic rod has length 5 cm and diameter 1 cm. It has uniform magnetization \( 5 \times 10^3 \, \text{A/m} \). Its net magnetic dipole moment is nearly

• A. \( 2.5 \times 10^{-2} \, \frac{\text{J}}{\text{T}} \)
• B. \( 0.5 \times 10^{-2} \, \frac{\text{J}}{\text{T}} \)
• C. \( 2 \times 10^{-2} \, \frac{\text{J}}{\text{T}} \)
• D. \( 10 \times 10^{-2} \, \frac{\text{J}}{\text{T}} \)

Hint:

The magnetic dipole moment of a magnetized body depends on its magnetization and volume. For a cylindrical rod, use the formula \( M = M_0 \times \pi r^2 L \).

Answer 1

The Correct Option is C

Step 1: Formula for magnetic dipole moment.
The magnetic dipole moment of a rod is given by: \[ M = \text{magnetization} \times \text{volume} = M_0 \times (\pi r^2 L) \] where \( M_0 \) is the magnetization, \( r \) is the radius, and \( L \) is the length of the rod.
Step 2: Substituting the values.
Given, - Length \( L = 5 \, \text{cm} \), - Diameter \( 1 \, \text{cm} \) (so radius \( r = 0.5 \, \text{cm} \)), - Magnetization \( M_0 = 5 \times 10^3 \, \text{A/m} \). The magnetic dipole moment is: \[ M = (5 \times 10^3) \times \left( \pi (0.5)^2 \times 5 \times 10^{-2} \right) = 2 \times 10^{-2} \, \frac{\text{J}}{\text{T}} \] Step 3: Conclusion.
The correct answer is (C), \( 2 \times 10^{-2} \, \frac{\text{J}}{\text{T}} \).