A magnetic needle lying parallel to a magneti | vedclass

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

Question

(A) The work done $W$ in rotating a magnetic dipole of magnetic moment $M$ in a magnetic field $B$ from an angle $	heta_1 = 0^{\circ}$ to $	heta_2 =

Vedclass · Accepted Answer

$\sqrt{3} W$

Vedclass · Answer

(A) The work done $W$ in rotating a magnetic dipole of magnetic moment $M$ in a magnetic field $B$ from an angle $	heta_1 = 0^{\circ}$ to $	heta_2 = 60^{\circ}$ is given by:$W = MB(1 - \cos 60^{\circ}) = MB(1 - 0.5) = \frac{MB}{2}$.From this,we find $MB = 2W$.The torque $	au$ required to maintain the needle at $	heta = 60^{\circ}$ is given by:$	au = MB \sin 60^{\circ}$.Substituting $MB = 2W$ and $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$:$	au = (2W) 	imes \frac{\sqrt{3}}{2} = \sqrt{3} W$.

$A$ magnetic needle lying parallel to a magnetic field is turned through $60^{\circ}$. The work done on it is $W$. The torque required to maintain the magnetic needle in the position mentioned above is

Similar Questions

The major product of the reaction is :

If the extremities of a diagonal of a square are $(1, -2, 3)$ and $(2, -3, 5)$,then the length of its side is

Three rods of same dimensions have thermal conductivities $3K$,$2K$,and $K$. They are arranged as shown in the figure. In the steady state,the temperature of the junction $P$ is

The slope of the normal to the curve $x = \sqrt{t}$ and $y = t - \frac{1}{\sqrt{t}}$ at $t = 4$ is ...

Which one of the following is an isotonic pair of solutions?

Vedclass Test Series

Exam Paper Generator

Online Exam Module