$A$ loop of irregular shape made of flexible conducting wire carrying a clockwise current is placed in a uniform inward magnetic field,such that its plane is perpendicular to the field. Then the loop:

  • A
    Experiences a magnetic force.
  • B
    Develops an induced current for a short time.
  • C
    Changes to a circular shape.
  • D
    All of these.

Explore More

Similar Questions

Match the following and find the correct pairs.
List-$I$List-$II$
$(A)$ Fleming's left hand rule$(i)$ Direction of induced current
$(B)$ Right hand thumb rule(ii) Magnitude and direction of magnetic induction
$(C)$ Biot-Savart law(iii) Direction of force due to magnetic induction
$(D)$ Fleming's right hand rule(iv) Direction of magnetic lines due to current

Which of the following is not a unit of magnetic induction?

$A$ charged particle of mass $m$ and charge $q$ moving under the influence of a uniform electric field $E\hat{i}$ and a uniform magnetic field $B\hat{k}$ follows a trajectory from point $P$ to $Q$ as shown in the figure. The velocities at $P$ and $Q$ are respectively $v\hat{i}$ and $-2v\hat{j}$. Which of the following statements $(A, B, C, D)$ are correct? (Trajectory shown is schematic and not to scale)
$(A)$ $E = \frac{3}{4}\left(\frac{mv^{2}}{qa}\right)$
$(B)$ Rate of work done by the electric field at $P$ is $\frac{3}{4}\left(\frac{mv^{3}}{a}\right)$
$(C)$ Rate of work done by both the fields at $Q$ is zero
$(D)$ The difference between the magnitude of angular momentum of the particle at $P$ and $Q$ is $2mav$.

$A$ light charged particle is revolving in a circle of radius $r$ due to the electrostatic attraction of a static heavy particle with an opposite charge. How does the magnetic field $B$ at the centre of the circle,produced by the moving charge,depend on $r$?

An electron is projected with velocity $v_0$ in a uniform electric field $E$ perpendicular to the field. Again,it is projected with velocity $v_0$ perpendicular to a uniform magnetic field $B$. If $r_1$ is the initial radius of curvature just after entering the electric field and $r_2$ is the initial radius of curvature just after entering the magnetic field,then the ratio $r_1:r_2$ is equal to:

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