(A) For the wire to reach terminal velocity,its acceleration must be zero.At terminal velocity,the magnetic force acting upwards must balance the grav

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(A) For the wire to reach terminal velocity,its acceleration must be zero.At terminal velocity,the magnetic force acting upwards must balance the gravitational force acting downwards.$F_m = mg$$I l B = mg$Since the induced electromotive force $(EMF)$ is $\varepsilon = B l v_0$,where $v_0$ is the terminal velocity,the induced current $I$ is given by:$I = \frac{\varepsilon}{R} = \frac{B l v_0}{R}$Substituting this into the force balance equation:$\left( \frac{B l v_0}{R} \right) l B = mg$$\frac{B^2 l^2 v_0}{R} = mg$Solving for $v_0$:$v_0 = \frac{mgR}{B^2 l^2}$

Class 12 Physics Electromagnetic Induction Motional EMI (Induced Parameter)

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$A$ horizontal wire is free to slide on the vertical rails of a conducting frame as shown in the figure. The wire has a mass $m$ and length $l$,and the resistance of the circuit is $R$. If a uniform magnetic field $B$ is directed perpendicular to the frame,find the terminal speed of the wire as it falls under the force of gravity.

A
$\frac{mgR}{B^2l^2}$
B
$\frac{mg}{Bl}$
C
$\frac{2mgR}{B^2l^2}$
D
$0$

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Electromagnetic Induction

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Motional EMI (Induced Parameter)

$A$ conducting rod $PQ$ of length $L = 1.0\, m$ is moving with a uniform speed $v = 20\, m/s$ in a uniform magnetic field $B = 4.0\, T$ directed into the paper. $A$ capacitor of capacity $C = 10\, \mu F$ is connected as shown in the figure. Then:

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When a rod of length $l$ is rotated with angular velocity $\omega$ in a uniform magnetic field of induction $B$ perpendicular to the plane of rotation,about one of its ends,the induced emf across its ends is:

MediumMHT CET 2010

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An aeroplane,with its wings spread $10 \, m$,is flying at a speed of $180 \, km/h$ in a horizontal direction. The total intensity of the Earth's magnetic field at that location is $2.5 \times 10^{-4} \, Wb/m^2$ and the angle of dip is $60^{\circ}$. The $EMF$ induced between the tips of the plane's wings will be ...... $mV$.

MediumJEE MAIN 2021

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$A$ constant force $F$ is applied to a conducting rod of length $l$ moving with constant speed $V$ on two parallel conducting rails connected at the ends by a resistance $R$ in a uniform magnetic field $B$,as shown. If the current flowing through the circuit is $I$,then:

Medium

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Explain the motional $emf$ by the Lorentz force acting on the free charge carriers of a conductor.

Medium

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$A$ conducting rod $PQ$ of length $L = 1.0\, m$ is moving with a uniform speed $v = 20\, m/s$ in a uniform magnetic field $B = 4.0\, T$ directed into the paper. $A$ capacitor of capacity $C = 10\, \mu F$ is connected as shown in the figure. Then:

When a rod of length $l$ is rotated with angular velocity $\omega$ in a uniform magnetic field of induction $B$ perpendicular to the plane of rotation,about one of its ends,the induced emf across its ends is:

$A$ constant force $F$ is applied to a conducting rod of length $l$ moving with constant speed $V$ on two parallel conducting rails connected at the ends by a resistance $R$ in a uniform magnetic field $B$,as shown. If the current flowing through the circuit is $I$,then:

Explain the motional $emf$ by the Lorentz force acting on the free charge carriers of a conductor.

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