A circular coil of radius 2 text{ cm} and 125 | vedclass

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(B) The torque $	au$ acting on a current-carrying coil in a magnetic field is given by the formula $	au = NIAB \sin 	heta$, where $N$ is the number

Vedclass · Accepted Answer

$314$

Vedclass · Answer

(B) The torque $	au$ acting on a current-carrying coil in a magnetic field is given by the formula $	au = NIAB \sin 	heta$, where $N$ is the number of turns, $I$ is the current, $A$ is the area of the coil, $B$ is the magnetic field, and $	heta$ is the angle between the normal to the coil (axis) and the magnetic field.Given: $N = 125$, $I = 1 	ext{ A}$, $r = 2 	ext{ cm} = 0.02 	ext{ m}$, $B = 0.4 	ext{ T}$, and $	heta = 30^{\circ}$.First, calculate the area $A = \pi r^2 = \pi (0.02)^2 = 4\pi 	imes 10^{-4} 	ext{ m}^2$.Now, substitute the values into the torque formula:$	au = 125 	imes 1 	imes (4\pi 	imes 10^{-4}) 	imes 0.4 	imes \sin(30^{\circ})$$	au = 125 	imes 4 	imes 3.14 	imes 10^{-4} 	imes 0.4 	imes 0.5$$	au = 500 	imes 3.14 	imes 10^{-4} 	imes 0.2$$	au = 100 	imes 3.14 	imes 10^{-4} = 314 	imes 10^{-4} 	ext{ N.m}$.Comparing this with $\alpha 	imes 10^{-4} 	ext{ N.m}$, we get $\alpha = 314$.

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