A closely wound solenoid of 2000 turns and area of cross-section $1.5 imes 10^{-4} ext{ m}^2$ carries a cu

Question

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

Accepted Answer

1. **Magnetic Moment:** The solenoid acts like a magnetic dipole. Its magnetic moment ($m$) is given by the product of the number of turns ($N$), the current ($I$), and the cross-sectional area ($A$): $m = NIA$ .
   $$m = 2000 	imes 2.0 	imes 1.5 	imes 10^{-4} = 0.6 	ext{ A m}^2$$
2. **Torque Calculation:** The torque ($	au$) exerted on a magnetic dipole in a uniform magnetic field ($B$) is given by the cross product $\vec{	au} = \vec{m} 	imes \vec{B}$, with magnitude $	au = mB \sin	heta$, where $	heta$ is the angle between the magnetic moment (axis of the solenoid) and the magnetic field .
3. **Substitution:** Given $	heta = 30^{\circ}$ and $B = 5 	imes 10^{-2} 	ext{ T}$:
   $$	au = 0.6 	imes (5 	imes 10^{-2}) 	imes \sin(30^{\circ})$$
   $$	au = 0.6 	imes 0.05 	imes 0.5$$
   $$	au = 1.5 	imes 10^{-2} 	ext{ N m}$$

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