A 250-turn rectangular coil with a length of $2.1 ext{ cm}$ and a width of $1.25 ext{ cm}$ carries a curre

Question

A 250-turn rectangular coil with a length of $2.1 	ext{ cm}$ and a width of $1.25 	ext{ cm}$ carries a current of $85 	ext{ }\mu	ext{A}$ and is subjected to a magnetic field of strength $0.85 	ext{ T}$. What is the work done to rotate the coil by $180^\circ$ against the torque?

Accepted Answer

1. **Magnetic Moment ($m$):** The current-carrying coil behaves as a magnetic dipole. The magnetic moment is given by $m = NIA$, where $N$ is the number of turns, $I$ is the current, and $A$ is the area of the coil.
   - $N = 250$
   - $I = 85 	ext{ }\mu	ext{A} = 85 	imes 10^{-6} 	ext{ A}$
   - $A = 	ext{length} 	imes 	ext{width} = (2.1 	imes 10^{-2} 	ext{ m}) 	imes (1.25 	imes 10^{-2} 	ext{ m}) \approx 2.625 	imes 10^{-4} 	ext{ m}^2$
   $$ m = 250 	imes (85 	imes 10^{-6}) 	imes (2.625 	imes 10^{-4}) \approx 5.58 	imes 10^{-6} 	ext{ A m}^2 $$
2. **Work Done ($W$):** The work done to rotate a magnetic dipole in a uniform magnetic field $B$ from an initial angle $	heta_1$ to a final angle $	heta_2$ is equal to the change in potential energy ($U = -mB \cos 	heta$). Rotating 'against the torque' typically implies moving from the stable equilibrium ($	heta_1 = 0^\circ$) to the unstable equilibrium ($	heta_2 = 180^\circ$).
   $$ W = U_f - U_i = -mB(\cos 180^\circ - \cos 0^\circ) $$
   $$ W = -mB(-1 - 1) = 2mB $$
   
3. **Calculation:**
   $$ W = 2 	imes (5.58 	imes 10^{-6}) 	imes 0.85 $$
   $$ W \approx 9.48 	imes 10^{-6} 	ext{ J} = 9.48 	ext{ }\mu	ext{J} $$
   The calculated value ($9.48 	ext{ }\mu	ext{J}$) is closest to the option $9.1 	ext{ }\mu	ext{J}$. (Discrepancy likely due to rounding of area dimensions in the original question source).

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