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Step-by-Step Solution
Step 1: Identify the Physical Quantities
• The coil has N turns and each turn has a radius $r$.
• The current flowing in the coil is $I$.
• The magnetic field is $\vec{B} = B \hat{i}$ (along the x-axis).
• The coil lies in the XZ-plane. Hence, its area vector is along the y-axis.
Step 2: Write Down the Expression for Magnetic Moment
When a coil of N turns carries a current $I$, the magnitude of its magnetic moment $\vec{m}$ is given by:
$$
m = N \times I \times A,
$$
where $A$ is the area of the coil. For a circular loop of radius $r$:
$$
A = \pi r^2.
$$
Therefore,
$$
m = N \, I \, \pi r^2.
$$
The direction of $\vec{m}$ is perpendicular to the plane of the coil (along the y-axis in this setup).
Step 3: Determine the Torque
The torque on a current-carrying loop placed in a uniform magnetic field $\vec{B}$ is given by the vector product:
$$
\vec{\tau} = \vec{m} \times \vec{B}.
$$
In this problem, $\vec{m}$ lies along the y-axis and $\vec{B}$ is along the x-axis. The angle between them is $90^\circ$, so the magnitude of the torque is:
$$
\tau = m B \sin(90^\circ) = m B = (N I \pi r^2) \, B.
$$
Step 4: Final Expression for the Torque
Substituting $m = N I \pi r^2$:
$$
\tau = N I \pi r^2 \, B.
$$
This matches the given correct answer:
$$
\tau = B \,\pi r^2\, I \, N.
$$
Correct Answer: $ B \,\pi r^2\, I \, N $
