(D) The magnetic flux linked with the coil is given by $\phi = nBA \cos \theta$.Initially,the axis of the coil is parallel to the magnetic field,so $\

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(D) The magnetic flux linked with the coil is given by $\phi = nBA \cos \theta$.Initially,the axis of the coil is parallel to the magnetic field,so $\theta_1 = 0^o$ and $\phi_1 = nBA \cos 0^o = nBA$.After rotating the coil by $180^o$ about its diameter,the new angle is $\theta_2 = 180^o$,so $\phi_2 = nBA \cos 180^o = -nBA$.The change in magnetic flux is $\Delta \phi = \phi_2 - \phi_1 = -nBA - nBA = -2nBA$.The induced charge $Q$ is given by $Q = \frac{|\Delta \phi|}{R} = \frac{2nBA}{R}$.Rearranging the formula to solve for the magnetic induction $B$,we get $B = \frac{QR}{2nA}$.

Class 12 Physics Electromagnetic Induction Motional EMI (Induced Parameter)

Medium

$A$ small coil is introduced between the poles of an electromagnet so that its axis coincides with the magnetic field direction. The number of turns is $n$ and the cross-sectional area of the coil is $A$. When the coil turns through $180^o$ about its diameter,the charge flowing through the coil is $Q$. The total resistance of the circuit is $R$. What is the magnitude of the magnetic induction?

A
$\frac{QR}{nA}$
B
$\frac{2QR}{nA}$
C
$\frac{Qn}{2RA}$
D
$\frac{QR}{2nA}$

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Class 12

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Electromagnetic Induction

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Motional EMI (Induced Parameter)

$A$ horizontal straight wire $20 \; m$ long extending from east to west is falling with a speed of $5.0 \; m/s$ at right angles to the horizontal component of the earth's magnetic field $0.30 \times 10^{-4} \; Wb/m^2$. The instantaneous value of the emf induced in the wire will be ......... $mV$.

MediumAIEEE 2011

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$A$ simple pendulum with a bob of mass $m$ and a conducting wire of length $L$ swings under gravity through an angle $2\theta$. The Earth's magnetic field component in the direction perpendicular to the swing is $B$. The maximum potential difference induced across the pendulum is

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$A$ long solenoid with $2000$ turns per meter has a small loop of radius $3 \,cm$ placed inside the solenoid normal to its axis. If the current through the solenoid increases steadily from $1.5 \,A$ to $5.5 \,A$ in $\frac{\pi^2}{100} \,s$, the induced emf in the loop is (in $\,mV$)

MediumAP EAMCET 2019

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If we double the velocity of a rod moving in a uniform magnetic field, then the induced $emf$ will increase by how many times?

Easy

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$A$ thin semicircular conducting ring of radius $R$ is falling with its plane vertical in a horizontal magnetic induction $B$. At the position $MNQ$,the speed of the ring is $V$ and the potential difference developed across the ring is

MediumIIT 1996

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$A$ horizontal straight wire $20 \; m$ long extending from east to west is falling with a speed of $5.0 \; m/s$ at right angles to the horizontal component of the earth's magnetic field $0.30 \times 10^{-4} \; Wb/m^2$. The instantaneous value of the emf induced in the wire will be ......... $mV$.

$A$ simple pendulum with a bob of mass $m$ and a conducting wire of length $L$ swings under gravity through an angle $2\theta$. The Earth's magnetic field component in the direction perpendicular to the swing is $B$. The maximum potential difference induced across the pendulum is

$A$ long solenoid with $2000$ turns per meter has a small loop of radius $3 \,cm$ placed inside the solenoid normal to its axis. If the current through the solenoid increases steadily from $1.5 \,A$ to $5.5 \,A$ in $\frac{\pi^2}{100} \,s$, the induced emf in the loop is (in $\,mV$)

If we double the velocity of a rod moving in a uniform magnetic field, then the induced $emf$ will increase by how many times?

$A$ thin semicircular conducting ring of radius $R$ is falling with its plane vertical in a horizontal magnetic induction $B$. At the position $MNQ$,the speed of the ring is $V$ and the potential difference developed across the ring is

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