Consider a current-carrying wire as shown in the figure. If the radius of the curved part of the wire is $R$ and the linear parts are assumed to be very long,then the magnetic induction of the field at the point $O$ is

  • A
    $\frac{\mu_0}{4 \pi} \frac{i}{R}(2+\pi)$
  • B
    $\frac{\mu_0}{2 \pi R}$
  • C
    $\frac{\mu_0}{2} \frac{i}{R}$
  • D
    $\frac{\mu_0}{4} \frac{i}{R}$

Explore More

Similar Questions

$A$ particle of specific charge $(q/m)$ is projected from the origin of coordinates with initial velocity $(u\hat{i} - v\hat{j})$. Uniform electric and magnetic fields exist in the region along the $+y$ direction,of magnitude $E$ and $B$ respectively. The particle will definitely return to the origin once if:

The figure shows a circular loop of radius $a$ with two long parallel wires (numbered $1$ and $2$) all in the plane of the paper. The distance of each wire from the centre of the loop is $d$. The loop and the wires are carrying the same current $I$. The current in the loop is in the counterclockwise direction if seen from above.
$1.$ When $d \approx a$ but wires are not touching the loop,it is found that the net magnetic field on the axis of the loop is zero at a height $h$ above the loop. In that case
$(A)$ current in wire $1$ and wire $2$ is in the direction $PQ$ and $RS$,respectively and $h \approx a$
$(B)$ current in wire $1$ and wire $2$ is in the direction $PQ$ and $SR$,respectively and $h \approx a$
$(C)$ current in wire $1$ and wire $2$ is in the direction $PQ$ and $SR$,respectively and $h \approx 1.2 a$
$(D)$ current in wire $1$ and wire $2$ is in the direction $PQ$ and $RS$,respectively and $h \approx 1.2 a$
$2.$ Consider $d \gg a$,and the loop is rotated about its diameter parallel to the wires by $30^{\circ}$ from the position shown in the figure. If the currents in the wires are in the opposite directions,the torque on the loop at its new position will be (assume that the net field due to the wires is constant over the loop)
$(A)$ $\frac{\mu_0 I^2 a^2}{d}$ $(B)$ $\frac{\mu_0 I^2 a^2}{2 d}$ $(C)$ $\frac{\sqrt{3} \mu_0 I^2 a^2}{d}$ $(D)$ $\frac{\sqrt{3} \mu_0 I^2 a^2}{2 d}$
Give the answer for question $1$ and $2$.

When a proton is released from rest in a room,it starts with an initial acceleration $a_0$ towards west. When it is projected towards north with a speed $v_0$,it moves with an initial acceleration $3a_0$ towards west. The electric and magnetic fields in the room are

Choose the correct statement.

The net magnetic field at the centre $O$ of the circle due to the current-carrying loop as shown in the figure is $(\theta < 180^\circ)$.

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