Question

If the magnetization of a bar magnet of area of cross-section 0.75 cm$^2$ and magnetic moment 2.7 Am$^2$ is $4 \times 10^5$ Am$^{-1}$, then the length of the magnet is

• A. 15 cm
• B. 9 cm
• C. 6 cm
• D. 12 cm

Hint:

- Formula: Magnetic moment = Magnetization × Volume.
- Convert units carefully: cm$^2$ → m$^2$, Am$^{-1}$ stays as is.
- Cross-check: $M_m = I \cdot A \cdot l$. Rearrange to find $l$.
- Remember that magnetization is the magnetic moment per unit volume.
- Practice similar problems for different cross-sections and magnetizations to strengthen understanding.

Answer 1

The Correct Option is B

1. Magnetic moment of a bar magnet: $M_m = \text{magnetization} \times \text{volume} = I \cdot V$.
2. Volume $V = A \cdot l$, where $A$ = area of cross-section, $l$ = length.
3. Given: $M_m = 2.7$ Am$^2$, $I = 4 \times 10^5$ Am$^{-1}$, $A = 0.75$ cm$^2 = 0.75 \times 10^{-4}$ m$^2$.
4. Length: $l = \frac{M_m}{I \cdot A} = \frac{2.7}{4 \times 10^5 \cdot 0.75 \times 10^{-4}} = \frac{2.7}{30} = 0.09$ m = 9 cm