(C) The motional electromotive force $(EMF)$ induced in the rod is given by $\varepsilon = B v L$.Given: $B = 0.5 \, T$,$v = 0.2 \, m/s$,$L = 0.1 \, m

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$q_A = +0.2 \, \mu C, q_B = -0.2 \, \mu C$

(C) The motional electromotive force $(EMF)$ induced in the rod is given by $\varepsilon = B v L$.Given: $B = 0.5 \, T$,$v = 0.2 \, m/s$,$L = 0.1 \, m$.$\varepsilon = (0.5 \, T) \times (0.2 \, m/s) \times (0.1 \, m) = 0.01 \, V$.Using the right-hand rule for the motional $EMF$,the upper end of the rod becomes positive and the lower end becomes negative. Consequently,plate $A$ of the capacitor becomes positively charged and plate $B$ becomes negatively charged.The charge on the capacitor is $q = C \varepsilon$.$q = (20 \times 10^{-6} \, F) \times (0.01 \, V) = 0.2 \times 10^{-6} \, C = 0.2 \, \mu C$.Thus,$q_A = +0.2 \, \mu C$ and $q_B = -0.2 \, \mu C$.

Class 12 Physics Electromagnetic Induction Motional EMI (Induced Parameter)

Medium

$A$ conducting rod of length $L = 0.1 \, m$ is moving with a uniform speed $v = 0.2 \, m/s$ on conducting rails in a magnetic field $B = 0.5 \, T$ as shown. On one side,the end of the rails is connected to a capacitor of capacitance $C = 20 \, \mu F$. Then the charges on the capacitor plates $A$ and $B$ are:

A
$q_A = 0, q_B = 0$
B
$q_A = +20 \, \mu C, q_B = -20 \, \mu C$
C
$q_A = +0.2 \, \mu C, q_B = -0.2 \, \mu C$
D
$q_A = -0.2 \, \mu C, q_B = +0.2 \, \mu C$

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Electromagnetic Induction

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Motional EMI (Induced Parameter)

$A$ circular coil of radius $30 \ cm$,having $1$ turn and a resistance of ${\pi ^2} \ \Omega$,is placed in a magnetic field of $10^{-2} \ T$. The coil rotates about its diameter with a speed of $200 \ rpm$ in a direction perpendicular to the magnetic field. The value of the induced $AC$ current in the coil will be . . . . . . $mA$.

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$A$ conductor of length $l$ and mass $m$ can slide along a pair of vertical metal guides connected by a resistance $R$,as shown in the figure. Friction,resistance of the conductor,and guide rails are negligible. There exists a horizontal uniform magnetic field of strength $B$ normal to the plane of the page and directed outward. The terminal speed of fall under the influence of gravity is:

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The magnetic field in a region is given by $\vec B = B_0(1 + \frac{x}{a})\hat k$. $A$ square loop of edge length $d$ is placed with its edges along the $x$ and $y$ axes. The loop is moved with a constant velocity $\vec V = V_0\hat i$. The $emf$ induced in the loop is:

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$A$ copper rod $AB$ of length $L$,pivoted at one end $A$,rotates at a constant angular velocity $\omega$,at right angles to a uniform magnetic field of induction $B$. The e.m.f. developed between the midpoint $C$ of the rod and end $B$ is

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$A$ conducting rod $PQ$ of length $L = 1.0\, m$ is moving with 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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The magnetic field in a region is given by $\vec B = B_0(1 + \frac{x}{a})\hat k$. $A$ square loop of edge length $d$ is placed with its edges along the $x$ and $y$ axes. The loop is moved with a constant velocity $\vec V = V_0\hat i$. The $emf$ induced in the loop is:

$A$ copper rod $AB$ of length $L$,pivoted at one end $A$,rotates at a constant angular velocity $\omega$,at right angles to a uniform magnetic field of induction $B$. The e.m.f. developed between the midpoint $C$ of the rod and end $B$ is

$A$ conducting rod $PQ$ of length $L = 1.0\, m$ is moving with 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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