A 250 turn rectangular coil of length $2.1 ext{ cm}$ and width $1.25 ext{ cm}$ carries a current of $85 \,

Question

A 250 turn rectangular coil of length $2.1 	ext{ cm}$ and width $1.25 	ext{ cm}$ carries a current of $85 \, \mu	ext{A}$ and subjected to the magnetic field of strength $0.85 	ext{ T}$. Work done for rotating the coil by $180^{\circ}$ against the torque is:

Accepted Answer

1. **Magnetic Moment ($m$):** The rectangular coil acts as a magnetic dipole. Its magnetic moment is given by $m = NIA$, where $N$ is the number of turns, $I$ is the current, and $A$ is the area .
   - $N = 250$
   - $I = 85 \, \mu	ext{A} = 85 	imes 10^{-6} 	ext{ A}$
   - $A = 	ext{length} 	imes 	ext{width} = (2.1 	imes 10^{-2} 	ext{ m}) 	imes (1.25 	imes 10^{-2} 	ext{ m}) \approx 2.625 	imes 10^{-4} 	ext{ m}^2$
   - $m = 250 	imes (85 	imes 10^{-6}) 	imes (2.625 	imes 10^{-4}) \approx 5.58 	imes 10^{-6} 	ext{ A m}^2$
2. **Work Done ($W$):** The work done in rotating a magnetic dipole from an initial angle $	heta_1$ to a final angle $	heta_2$ in a uniform magnetic field $B$ is given by the change in potential energy: $W = \Delta U = mB(\cos	heta_1 - \cos	heta_2)$ .
   - "Rotating by $180^{\circ}$ against the torque" typically implies rotating from the stable equilibrium position ($	heta_1 = 0^{\circ}$) to the unstable equilibrium position ($	heta_2 = 180^{\circ}$).
   - $W = mB(\cos 0^{\circ} - \cos 180^{\circ}) = mB(1 - (-1)) = 2mB$
3. **Calculation:**
   - $W = 2 	imes (5.58 	imes 10^{-6}) 	imes 0.85$
   - $W \approx 9.48 	imes 10^{-6} 	ext{ J} = 9.48 \, \mu	ext{J}$
4. **Conclusion:** The value closest to the calculated work is $9.4 \, \mu	ext{J}$.

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