$A$ wire $ab$ of length $l$,mass $m$,and resistance $R$ slides on a smooth,thick pair of metallic rails joined at the bottom as shown in the figure. The plane of the rails makes an angle $\theta$ with the horizontal. $A$ vertical magnetic field $B$ exists in the region. If the wire slides on the rails at a constant speed $v$,then which of the following can be correct for $B$?

  • A
    $B = \sqrt {\frac{{mgR}}{{v{l^2}\,\cos\, \theta }}} $
  • B
    $B = \sqrt {\frac{{mgR\,\sin \,\theta }}{{v{l^2}\,\cos \,\theta }}} $
  • C
    $B = \sqrt {\frac{{mgR\,\sin \,\theta }}{{v{l^2}\,{{\cos }^2}\,\theta }}} $
  • D
    $B = \sqrt {\frac{{mgR\,\sin \,\theta }}{{v{l^2}}}}$

Explore More

Similar Questions

The figure shows a metal rod $PQ$ resting on the smooth rails $AB$ and positioned between the poles of a permanent magnet. The rails,the rod,and the magnetic field are in three mutually perpendicular directions. $A$ galvanometer $G$ connects the rails through a switch $K$. Length of the rod $= 15 \; cm$,$B = 0.50 \; T$,resistance of the closed loop containing the rod $= 9.0 \; m\Omega$. Assume the field to be uniform.
$(a)$ Suppose $K$ is open and the rod is moved with a speed of $12 \; cm \; s^{-1}$ in the direction shown. Give the polarity and magnitude of the induced $emf$.
$(b)$ Is there an excess charge built up at the ends of the rod when $K$ is open? What if $K$ is closed?
$(c)$ With $K$ open and the rod moving uniformly,there is no net force on the electrons in the rod $PQ$ even though they do experience magnetic force due to the motion of the rod. Explain.
$(d)$ What is the retarding force on the rod when $K$ is closed?
$(e)$ How much power is required (by an external agent) to keep the rod moving at the same speed $(= 12 \; cm \; s^{-1})$ when $K$ is closed? How much power is required when $K$ is open?
$(f)$ How much power is dissipated as heat in the closed circuit? What is the source of this power?
$(g)$ What is the induced $emf$ in the moving rod if the magnetic field is parallel to the rails instead of being perpendicular?

$A$ simple pendulum made of a mass of $10 \ g$ and a metallic wire of length $10 \ cm$ is suspended vertically in a uniform magnetic field of $2 \ T$. The magnetic field direction is perpendicular to the plane of oscillations of the pendulum. If the pendulum is released from an angle of $60^{\circ}$ with the vertical, then the maximum induced $EMF$ between the point of suspension and the point of oscillation is . . . . . . $mV$. (Take $g = 10 \ m/s^2$)

$A$ rectangular loop of wire is placed in the $XY$-plane with its side of length $3 \,cm$ parallel to the $X$-axis and the side of length $4 \,cm$ parallel to the $Y$-axis. It is moving in the positive $X$-direction with the speed $10 \,cm/s$. $A$ magnetic field exists in the space with its direction parallel to the $Z$-axis. The field decreases by $2 \times 10^{-3} \,T/cm$ along the positive $X$-axis and increases in time by $2 \times 10^{-2} \,T/s$. The induced emf in the wire is

$A$ metal rod of length '$L$' completes the circuit as shown. The area of the circuit is perpendicular to the magnetic field '$B$'. The total resistance of the circuit is '$R$'. The force needed to move the rod in the direction as shown with a constant speed '$V$' is

$A$ conductor $10 \ cm$ long is moved with a speed $1 \ m/s$ perpendicular to a magnetic field of strength $1000 \ A/m$. The e.m.f. induced in the conductor is [Given : $\mu_0 = 4 \pi \times 10^{-7} \ Wb/Am$]

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