A circular coil of 30 turns and radius 8.0, c | vedclass

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

Question

(C) The torque $	au$ acting on a current-carrying coil in a magnetic field is given by $	au = N I A B \sin 	heta$,where $	heta$ is the angle betwe

Vedclass · Accepted Answer

$3.1$

Vedclass · Answer

(C) The torque $	au$ acting on a current-carrying coil in a magnetic field is given by $	au = N I A B \sin 	heta$,where $	heta$ is the angle between the normal to the coil and the magnetic field.Given:Number of turns $N = 30$Radius $r = 8.0\, cm = 0.08\, m$Current $I = 6.0\, A$Magnetic field $B = 1.0\, T$Angle $	heta = 60^o$Area $A = \pi r^2 = 3.14 	imes (0.08)^2 = 3.14 	imes 0.0064 = 0.020096\, m^2$Substituting the values:$	au = 30 	imes 6.0 	imes 0.020096 	imes 1.0 	imes \sin 60^o$$	au = 180 	imes 0.020096 	imes \frac{\sqrt{3}}{2}$$	au = 90 	imes 0.020096 	imes 1.732$$	au \approx 3.133\, Nm$Rounding to two significant figures,the magnitude of the counter torque is $3.1\, Nm$.

Similar Questions

The $(\tau - \theta)$ graph for a current-carrying coil in a magnetic field is:

$A$ current-carrying loop is free to turn in a uniform magnetic field. The loop will come into equilibrium when its plane is inclined at

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