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Step-by-Step Solution
Step 1: Identify the Relevant Formula
The work done in rotating a current-carrying coil from an initial angle
$ \theta_{1} $ to a final angle $ \theta_{2} $ in a uniform magnetic field
$ B $ is given by the change in potential energy:
$$
W \;=\; M B (\cos \theta_{1} - \cos \theta_{2}),
$$
where $ M $ is the magnetic dipole moment of the coil,
$ M = NI A $ (with $N$ the number of turns, $I$ the current, and $A$ the area of the coil).
For a rotation through 180° ($ \theta_{2} - \theta_{1} = 180^\circ $), the expression reduces to
$ W = 2 M B $ because $ (\cos \theta_{1} - \cos \theta_{2}) = 2 $ when $ \theta_{2} - \theta_{1} = 180^\circ $.
Step 2: List the Given Data
Number of turns, $ N = 250 $
Length of the coil, $ l = 2.1 \text{ cm} = 2.1 \times 10^{-2} \text{ m} $
Width of the coil, $ w = 1.25 \text{ cm} = 1.25 \times 10^{-2} \text{ m} $
Current, $ I = 85 \,\mu\text{A} = 85 \times 10^{-6} \text{ A} $
Magnetic field, $ B = 0.85 \text{ T} $
Step 3: Calculate the Area of the Coil
The area of the rectangular coil is:
$$
A = l \times w
= (2.1 \times 10^{-2} \,\text{m}) \times (1.25 \times 10^{-2} \,\text{m})
= 2.625 \times 10^{-4} \,\text{m}^2.
$$
Step 4: Determine the Magnetic Dipole Moment $ M $
The magnetic dipole moment of the coil is given by:
$$
M = N I A.
$$
Substituting the values,
$$
M = 250 \times (85 \times 10^{-6} \,\text{A}) \times (2.625 \times 10^{-4} \,\text{m}^2).
$$
Let us perform this multiplication step by step:
$ 85 \times 10^{-6} = 8.5 \times 10^{-5} $
So,
$$
I A = (8.5 \times 10^{-5}) \times (2.625 \times 10^{-4})
= 2.23125 \times 10^{-8} \,\text{A} \cdot \text{m}^2.
$$
Multiply by $N = 250$:
$$
M = 250 \times 2.23125 \times 10^{-8}
= 5.578125 \times 10^{-6} \,\text{A} \cdot \text{m}^2.
$$
Step 5: Compute the Work Done in Rotating the Coil 180°
Using
$ W = 2 M B $:
$$
W = 2 \times \bigl(5.578125 \times 10^{-6}\bigr) \times 0.85 \,\text{J}.
$$
Numerically,
$$
W = 2 \times 5.578125 \times 10^{-6} \times 0.85
\approx 9.1 \times 10^{-6} \,\text{J}.
$$
Converting to microjoules ($\mu \text{J}$):
$$
9.1 \times 10^{-6} \,\text{J}
= 9.1 \,\mu \text{J}.
$$
Step 6: State the Final Answer
Therefore, the work done in rotating the coil by 180° is
$9.1 \,\mu \text{J}$.
