Question

A bar magnet has length 3 cm, cross-sectional area $2\ \text{cm}^2$ and magnetic moment $3\ \text{A m}^2$. The intensity of magnetisation of the bar magnet is

• A. $2 \times 10^5\ \text{A m}^{-1}$
• B. $3 \times 10^5\ \text{A m}^{-1}$
• C. $4 \times 10^5\ \text{A m}^{-1}$
• D. $5 \times 10^5\ \text{A m}^{-1}$

Hint:

An alternative equivalent definition for the intensity of magnetization is pole strength per unit cross-sectional area ($I = \frac{m_{\text{pole}}}{A}$). Both approaches yield identical results, but always make sure to convert metric prefixes like $\text{cm}^2$ carefully: remember that $1\ \text{cm}^2 = 10^{-4}\ \text{m}^2$, not $10^{-2}\ \text{m}^2$.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given the structural dimensions (length and cross-sectional area) along with the net magnetic dipole moment ($M$) of a uniform bar magnet. We need to calculate its interior intensity of magnetization ($I$).

Step 2: Key Formula or Approach:
The intensity of magnetization ($I$) is defined as the magnetic dipole moment per unit volume of the magnetic material:
$$I = \frac{M}{V}$$ The volume ($V$) of a uniform bar magnet can be calculated by multiplying its cross-sectional area ($A$) by its length ($L$):
$$V = A \cdot L$$

Step 3: Detailed Explanation:
Let's first convert all the given values into standard SI units:
Magnetic Dipole Moment, $M = 3\ \text{A m}^2$ Length of the magnet, $L = 3\ \text{cm} = 3 \times 10^{-2}\ \text{m}$ Cross-sectional area, $A = 2\ \text{cm}^2 = 2 \times (10^{-2}\ \text{m})^2 = 2 \times 10^{-4}\ \text{m}^2$ Next, compute the total geometric volume ($V$) occupied by the bar magnet:
$$V = A \cdot L = \left(2 \times 10^{-4}\ \text{m}^2\right) \times \left(3 \times 10^{-2}\ \text{m}\right) = 6 \times 10^{-6}\ \text{m}^3$$ Now, substitute the values for magnetic moment and volume into the intensity of magnetization formula:
$$I = \frac{M}{V} = \frac{3}{6 \times 10^{-6}}$$ Simplify the numerical fraction:
$$I = \frac{1}{2} \times 10^6 = 0.5 \times 10^6 = 5 \times 10^5\ \text{A m}^{-1}$$ This calculation gives an exact value of $5 \times 10^5\ \text{A m}^{-1}$ for the magnetization field intensity.

Step 4: Final Answer:
The intensity of magnetization of the bar magnet is $5 \times 10^5\ \text{A m}^{-1}$, which corresponds to option (D).