## Related questions with answers

A long piece of wire with a mass of 0.100 kg and a total length of 4.00 m is used to make a square coil with a side of 0.100 m. The coil is hinged along a horizontal side, carries a 3.40-A current, and is placed in a vertical magnetic field with a magnitude of 0.010 0 T. Find the torque acting on the coil due to the magnetic force at equilibrium.

Solutions

VerifiedIn this problem, a wire has mass $m = 0.100~\mathrm{kg}$ and length $l = 4.00~\mathrm{m}$. It makes a square coil of side $s = 0.100~\mathrm{m}$. It is hinged along a horizontal side, with current $I = 3.40~\mathrm{A}$. A vertical magnetic field is present with $B = 0.0100~\mathrm{T}$. For this part, we calculate the torque due to the magnetic force at equilibrium.

As mentioned in the solution of the proceeding part of the exercise, the magnetic torque will be given by

$\tau =NIBa^2\cos \theta ,$

where $N$ is the number of turns and was found to be 10, $I$ is the intensity, $B$ the magnetic field strength, $a$ the length of the side of the square, and $\theta$ the angle that the plane of the square makes with the vertical, which was found to be 3.97 $^{\circ}$. Thus, we can now simply substitute, finding

$\tau =10\cdot 3.4\cdot 0.01\cdot 0.1^2\cdot \cos 3.97^{\circ}=\boxed{3.39\cdot 10^{-3}~\mathrm{Nm}.}$

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