A coil in the shape of an equilateral triangl | vedclass

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(B) The torque $	au$ on a current-carrying coil in a magnetic field is given by $	au = \vec{m} 	imes \vec{B}$,where $\vec{m} = i \vec{A}$ is the ma

Vedclass · Accepted Answer

$2\left(\frac{	au}{\sqrt{3} B i}ight)^{1 / 2}$

Vedclass · Answer

(B) The torque $	au$ on a current-carrying coil in a magnetic field is given by $	au = \vec{m} 	imes \vec{B}$,where $\vec{m} = i \vec{A}$ is the magnetic moment.Since $\vec{B}$ is in the plane of the coil,the angle between the area vector $\vec{A}$ and the magnetic field $\vec{B}$ is $90^{\circ}$.Thus,the magnitude of the torque is $	au = i A B \sin 90^{\circ} = i A B$.The area $A$ of an equilateral triangle with side $l$ is $A = \frac{\sqrt{3}}{4} l^2$.Substituting this into the torque equation: $	au = i \left( \frac{\sqrt{3}}{4} l^2 ight) B$.Rearranging for $l^2$: $l^2 = \frac{4 	au}{\sqrt{3} i B}$.Taking the square root: $l = \sqrt{\frac{4 	au}{\sqrt{3} i B}} = 2 \left( \frac{	au}{\sqrt{3} i B} ight)^{1/2}$.

$A$ coil in the shape of an equilateral triangle of side $l$ is suspended between the pole pieces of a permanent magnet such that $\vec{B}$ is in the plane of the coil. If due to a current $i$ in the triangle a torque $\tau$ acts on it,the side $l$ of the triangle is

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