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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $\alpha = 314$.

Similar Questions

$A$ current-carrying loop is placed in a uniform magnetic field. The torque acting on the loop does not depend upon

$A$ coil having $100$ turns, area of $5 \times 10^{-3} \, m^2$, carrying current of $1 \, mA$ is placed in a uniform magnetic field of $0.20 \, T$ such that the plane of the coil is perpendicular to the magnetic field. The work done in turning the coil through $90^{\circ}$ is . . . . . . $\mu J$.

Current $I$ is carried in a wire of length $L$. If the wire is bent into a circular coil of single turn,the maximum torque in a given magnetic field $B$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module