$A$ rectangular loop has a sliding connector $PQ$ of length $2\ m$ and resistance $10\Omega$. It is moving with a speed $5\ m/s$ as shown. The set-up is placed in a uniform magnetic field $3\ T$ directed into the plane of the paper. Find the three currents $I_1$,$I_2$,and $I$.

  • A
    $I_1 = I_2 = 3\ A, I = 1\ A$
  • B
    $I_1 = I_2 = 5\ A, I = 2\ A$
  • C
    $I_1 = I_2 = 1\ A, I = 2\ A$
  • D
    $I_1 = I_2 = I = 2\ A$

Explore More

Similar Questions

$A$ conducting rod $AB$ of length $l$ is projected on a frictionless frame $PSRQ$ with velocity $v_0$ in a uniform magnetic field $\vec{B}$ directed into the plane. The velocity of the rod after time $t$ is

$A$ metallic rod of $1\; m$ length is rotated with a frequency of $50\; rev/s$,with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius $1\; m$,about an axis passing through the centre and perpendicular to the plane of the ring (Figure). $A$ constant and uniform magnetic field of $1\; T$ parallel to the axis is present everywhere. What is the $emf$ between the centre and the metallic ring?

$A$ horizontal wire of mass $m$,length $l$,and resistance $R$ is sliding on vertical rails in a uniform magnetic field $B$ directed perpendicularly. The terminal speed of the wire as it falls under the force of gravity is ($g =$ acceleration due to gravity).

Two long parallel horizontal rails,a distance $l$ apart,each having a resistance $\lambda$ per unit length,are joined at one end by a resistance $R$. $A$ perfectly conducting rod $MN$ of mass $m$ is free to slide along the rails without friction. There is a uniform magnetic field of induction $B$ normal to the plane of the paper and directed into the paper. $A$ variable force $F$ is applied to the rod $MN$ such that,as the rod moves,a constant current $i$ flows through the circuit. The applied force $F$ as a function of distance $x$ of the rod from $R$ is:

Difficult
View Solution

$A$ simple pendulum with a bob of mass $m$ and a conducting wire of length $L$ swings under gravity through an angle $\theta$. The component of the Earth's magnetic field in the direction perpendicular to the swing is $B$. The maximum e.m.f. induced across the pendulum is ($g=$ acceleration due to gravity).

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