The time period of a bar magnet suspended hor | vedclass

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

Question

(A) The time period $T$ of a bar magnet oscillating in a uniform magnetic field $B_H$ is given by the formula: $T = 2\pi \sqrt{\frac{I}{MB_H}}$.Here,$

Vedclass · Accepted Answer

Is directly proportional to the square root of its mass

Vedclass · Answer

(A) The time period $T$ of a bar magnet oscillating in a uniform magnetic field $B_H$ is given by the formula: $T = 2\pi \sqrt{\frac{I}{MB_H}}$.Here,$I$ is the moment of inertia of the bar magnet,$M$ is the magnetic moment,and $B_H$ is the horizontal component of the earth's magnetic field.The moment of inertia $I$ for a rectangular bar magnet of mass $m$,length $l$,and breadth $b$ is given by $I = \frac{m(l^2 + b^2)}{12}$.Substituting $I$ into the time period formula,we get $T = 2\pi \sqrt{\frac{m(l^2 + b^2)}{12MB_H}}$.Since $T \propto \sqrt{m}$,the time period is directly proportional to the square root of its mass.

The time period of a bar magnet suspended horizontally in the earth's magnetic field and allowed to oscillate:

Similar Questions

The radius of the coil of a Tangent galvanometer,which has $10$ turns,is $0.1\,m$. The current required to produce a deflection of $60^{\circ}$ $(B_H = 4 \times 10^{-5}\,T)$ is.....$A$

The tangent galvanometer,when connected in series with a standard resistance,can be used as:

Two bar magnets oscillate in a horizontal plane in Earth's magnetic field with time periods of $3\,s$ and $4\,s$ respectively. If their moments of inertia are in the ratio of $3: 2$,then the ratio of their magnetic moments will be.

At neutral temperature,the thermoelectric power $\left( \frac{dE}{dT} \right)$ has the value:

Vedclass Test Series

Exam Paper Generator

Online Exam Module