(B) The motional emf induced in the loop is given by: $V = B v l$.The current $i$ in the loop is given by Ohm's law: $i = \frac{V}{R} = \frac{B v l}{R

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(B) The motional emf induced in the loop is given by: $V = B v l$.The current $i$ in the loop is given by Ohm's law: $i = \frac{V}{R} = \frac{B v l}{R}$.As the loop falls,it experiences a magnetic force $F_m = i l B$ acting upwards. For the loop to fall with a constant terminal velocity $v$,the magnetic force must balance the gravitational force $mg$ acting downwards.Therefore,$i l B = m g$.Substituting the expression for $i$:$B \left( \frac{B v l}{R} \right) l = m g$$\frac{B^2 l^2 v}{R} = m g$$v = \frac{m g R}{B^2 l^2}$.

Class 12 Physics Electromagnetic Induction Motional EMI (Induced Parameter)

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$A$ long,rectangular conducting loop of width $l$,mass $m$,and resistance $R$ is placed partly in a perpendicular magnetic field $B$. It is pushed downwards with velocity $v$ so that it may continue to fall freely. The velocity $v$ is ($g=$ acceleration due to gravity).

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A
$\frac{m g R^2}{B l}$
B
$\frac{m g R}{B^2 l^2}$
C
$\frac{m g l}{B^2 R^2}$
D
$\frac{B^2 l^2 R}{m g}$

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Electromagnetic Induction

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Motional EMI (Induced Parameter)

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$A$ conducting rod of length $l$ moves with velocity $\upsilon$ in a direction parallel to a long wire carrying a steady current $I$. The axis of the rod is maintained perpendicular to the wire with the near end at a distance $r$ away as shown in the figure. Find the emf induced in the rod.

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$A$ conducting rod $AB$ moves parallel to the $X-$ axis in a uniform magnetic field,which is directed perpendicular to the plane of motion (outwards). The end $A$ of the rod gets:

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$A$ conducting rod $PQ$ of length $1 \ m$ is moving with a uniform speed $2 \ ms^{-1}$ in a uniform magnetic field of $4 \ T$ which is directed into the paper. $A$ capacitor of capacity $10 \ \mu F$ is connected as shown in the figure. Then,the charge on the plates of the capacitor are

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$A$ wheel with $10$ metallic spokes,each $0.5 \; m$ long,is rotated with a speed of $120 \; rev/min$ in a plane normal to the horizontal component of the Earth's magnetic field $H_{E}$ at a place. If $H_{E} = 0.4 \; G$ at the place,what is the induced $emf$ between the axle and the rim of the wheel? $(1 \; G = 10^{-4} \; T)$

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$A$ $10\,m$ long horizontal wire extends from North-East to South-West. It is falling with a speed of $5.0\,m/s$,at right angles to the horizontal component of the earth's magnetic field,which is $0.3 \times 10^{-4}\,Wb/m^2$. The value of the induced $emf$ in the wire is:

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$A$ long,rectangular conducting loop of width $l$,mass $m$,and resistance $R$ is placed partly in a perpendicular magnetic field $B$. It is pushed downwards with velocity $v$ so that it may continue to fall freely. The velocity $v$ is ($g=$ acceleration due to gravity).

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$A$ conducting rod of length $l$ moves with velocity $\upsilon$ in a direction parallel to a long wire carrying a steady current $I$. The axis of the rod is maintained perpendicular to the wire with the near end at a distance $r$ away as shown in the figure. Find the emf induced in the rod.

$A$ conducting rod $AB$ moves parallel to the $X-$ axis in a uniform magnetic field,which is directed perpendicular to the plane of motion (outwards). The end $A$ of the rod gets:

$A$ conducting rod $PQ$ of length $1 \ m$ is moving with a uniform speed $2 \ ms^{-1}$ in a uniform magnetic field of $4 \ T$ which is directed into the paper. $A$ capacitor of capacity $10 \ \mu F$ is connected as shown in the figure. Then,the charge on the plates of the capacitor are

$A$ $10\,m$ long horizontal wire extends from North-East to South-West. It is falling with a speed of $5.0\,m/s$,at right angles to the horizontal component of the earth's magnetic field,which is $0.3 \times 10^{-4}\,Wb/m^2$. The value of the induced $emf$ in the wire is:

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