$A$ circular coil of radius $2 \text{ cm}$ and $125 \text{ turns}$ carries a current of $1 \text{ A}$. The coil is placed in a uniform magnetic field of magnitude $0.4 \text{ T}$. The axis of the coil makes an angle of $30^{\circ}$ with the direction of the magnetic field. The torque acting on the coil is $\alpha \times 10^{-4} \text{ N.m}$. The value of $\alpha$ is . . . . . . .

  • A
    $218$
  • B
    $314$
  • C
    $428$
  • D
    $520$

Explore More

Similar Questions

$A$ 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 . . . . . . $\mu J$.

$A$ rectangular coil (dimension $5\,cm \times 2\,cm$) with $100\,turns$,carrying a current of $3\,A$ in the clockwise direction,is kept centered at the origin and in the $X-Z$ plane. $A$ magnetic field of $1\,T$ is applied along the $X$-axis. If the coil is tilted through $45^{\circ}$ about the $Z$-axis,then the torque on the coil is.....$Nm$.

$A$ thin wire of length $L$ made of an insulating material is bent to form a circular loop and a positive charge $q$ is given so that it is distributed uniformly around the circumference of the loop. The loop is then rotated with an angular speed $\omega$ about an axis passing through its centre. If a uniform magnetic field $B$ directed parallel to the plane of the loop is applied,then the magnitude of the magnetic torque on the loop is:

The torque required to hold a small circular coil of $10$ turns, area $1 \,mm^{2}$ and carrying a current of $\left(\frac{21}{44}\right) \,A$ in the middle of a long solenoid of $10^{3} \,turns/m$ carrying a current of $2.5 \,A$, with its axis perpendicular to the axis of the solenoid is

$A$ rectangular coil of length $0.12\, m$ and width $0.1\, m$ having $50$ turns of wire is suspended vertically in a uniform magnetic field of strength $0.2\, Wb/m^2$. The coil carries a current of $2\, A$. If the plane of the coil is inclined at an angle of $30^{\circ}$ with the direction of the field,the torque required to keep the coil in this position will be .......$Nm$.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo