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Question

(A) When a conducting rod of length $l$ moves with a velocity $V$ in a uniform magnetic field $B$ perpendicular to its length,an induced electromotive

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Current will flow from $A$ to $B$ through resistance.

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(A) When a conducting rod of length $l$ moves with a velocity $V$ in a uniform magnetic field $B$ perpendicular to its length,an induced electromotive force $(EMF)$ is generated across the rod given by $\varepsilon = VBl$.According to Fleming's Right-Hand Rule,the direction of the induced current is such that the upper end of the rod becomes at a higher potential and the lower end at a lower potential.Therefore,the current flows from the upper end to the lower end through the external circuit containing the resistance $R$. In the given diagram,the upper end is connected to $A$ and the lower end to $B$.Thus,the current flows from $A$ to $B$ through the resistance $R$.The potential difference across the resistance $R$ is equal to the induced $EMF$,which is $V = IR = VBl$.

$A$ 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:

Similar Questions

$A$ rod closing the circuit shown in the figure moves along a $U$-shaped wire at a constant speed $v$ under the action of a force $F$. The circuit is in a uniform magnetic field perpendicular to the plane. Calculate $F$ if the rate of heat generation in the circuit is $Q$.

As shown in the figure,a metal rod makes contact and completes the circuit. The circuit is perpendicular to the magnetic field with $B = 0.15 \, T$. If the resistance is $3 \, \Omega$,the force needed to move the rod as indicated with a constant speed of $2 \, m/s$ is:

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