Rankers Physics

Magnetic Properties of Matter: Practice Problem & Solution

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

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 moment of the solenoid is $M = N I A = 2000 \times 2.0 \times (1.5 \times 10^{-4}) = 0.6 J/T$. The torque acting on the solenoid is $\tau = M B \sin\theta = 0.6 \times (5 \times 10^{-2}) \times \sin(30^{\circ}) = 1.5 \times 10^{-2} Nm$.

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