CBSE · Class 12 · Physics · Magnetism and MatterA short bar magnet placed with its axis at $30^\circ$ with an external uniform magnetic field of $0.25 \text{ T}$ experiences a torque of magnitude equal to $4.5 \times 10^{-2} \text{ J}$. What is the magnitude of the magnetic moment of the magnet? If the length of the magnet is $5 \text{ cm}$, calculate its pole strength.

Given Data:

• Angle with magnetic field, $\theta = 30^\circ$
• External magnetic field, $B = 0.25 \text{ T}$
• Torque, $\tau = 4.5 \times 10^{-2} \text{ J}$ (or $\text{N}\cdot\text{m}$)
• Length of the magnet, $2l = 5 \text{ cm} = 5 \times 10^{-2} \text{ m}$

Step 1: Calculate the Magnetic Moment ($M$)\nThe formula for torque acting on a magnetic dipole in a uniform magnetic field is:

$$\tau = M B \sin\theta$$ \nRearranging the formula to solve for $M$: $$M = \frac{\tau}{B \sin\theta}$| \nSubstitute the given values into the equation: $$M = \frac{4.5 \times 10^{-2}}{0.25 \times \sin(30^\circ)}$$ \nSince $\sin(30^\circ) = 0.5$: $$M = \frac{4.5 \times 10^{-2}}{0.25 \times 0.5}$$ $$M = \frac{4.5 \times 10^{-2}}{0.125}$$ $$M = 36 \times 10^{-2} \text{ A}\cdot\text{m}^2 = 0.36 \text{ A}\cdot\text{m}^2$$

Step 2: Calculate the Pole Strength ($m$)\nThe magnetic moment is related to pole strength and length by the formula:

$$M = m \times 2l$$ \nRearranging for pole strength ($m$): $$m = \frac{M}{2l}$$ \nSubstitute the values of $M$ and $2l$: $$m = \frac{0.36 \text{ A}\cdot\text{m}^2}{5 \times 10^{-2} \text{ m}}$$ $$m = \frac{0.36}{0.05} = 7.2 \text{ A}\cdot\text{m}$$

Final Answer:

• The magnitude of the magnetic moment is $0.36 \text{ A}\cdot\text{m}^2$.
• The pole strength of the magnet is $7.2 \text{ A}\cdot\text{m}$.