Rankers Physics

Magnetic Properties of Matter: Practice Problem & Solution

5. A closely wound solenoid of 2000 turns and area of cross section $1.5 \times 10^{-4}\text{ m}^2$ carries a current of $2.0\text{ A}$. It is suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field $5 \times 10^{-2}\text{ tesla}$ making an angle of $30^\circ$ with the axis of the solenoid. The torque on the solenoid will be (2010 Mains)
$3 \times 10^{-3}\text{ Nm}$
$1.5 \times 10^{-3}\text{ Nm}$
$1.5 \times 10^{-2}\text{ Nm}$
$3 \times 10^{-2}\text{ Nm}$

Solution Explained:

To solve this problem, we apply the core principles of Magnetic Properties of Matter. Understanding the underlying formula is key to arriving at the correct answer below:

Magnetic dipole moment $M = NIA = 2000 \times 2.0 \times 1.5 \times 10^{-4} = 0.6\text{ A m}^2$.
Torque $\tau = MB \sin\theta = 0.6 \times (5 \times 10^{-2}) \times \sin 30^\circ = 1.5 \times 10^{-2}\text{ Nm}$.

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