(A) Given: Length $l = 10 \, cm = 0.1 \, m$,Magnetic field $B = 1.0 \, T$,Velocity $v = 20 \, cm/s = 0.2 \, m/s$,Angle $\theta = 60^\circ$.The motiona

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(A) Given: Length $l = 10 \, cm = 0.1 \, m$,Magnetic field $B = 1.0 \, T$,Velocity $v = 20 \, cm/s = 0.2 \, m/s$,Angle $\theta = 60^\circ$.The motional $emf$ induced in a conductor moving in a magnetic field is given by $E = B v l \sin \theta$,where $\theta$ is the angle between the velocity vector and the length of the wire.Substituting the values:$E = 1.0 \times 0.2 \times 0.1 \times \sin 60^\circ$$E = 0.02 \times \frac{\sqrt{3}}{2}$$E = 0.01 \times 1.732$$E = 0.01732 \, V$$E = 17.32 \times 10^{-3} \, V$.Thus,the induced $emf$ is approximately $17 \times 10^{-3} \, V$.

Class 12 Physics Electromagnetic Induction Motional EMI (Induced Parameter)

Medium

$A$ wire of length $10 \, cm$ translates in a direction making an angle of $60^\circ$ with its length. The plane of motion is perpendicular to a uniform magnetic field of $1.0 \, T$ that exists in the space. Find the $emf$ induced between the ends of the rod if the speed of translation is $20 \, cm/s$.

A
$17 \times 10^{-3} \, V$
B
$27 \times 10^{-3} \, V$
C
$7 \times 10^{-3} \, V$
D
$57 \times 10^{-3} \, V$

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Class 12

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Electromagnetic Induction

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Motional EMI (Induced Parameter)

Find the current in the wire for the configuration shown in the figure. Wire $PQ$ has negligible resistance. $\vec{B}$ is the magnetic field coming out of the paper. $\theta$ is a fixed angle made by $PQ$ as it travels smoothly over two conducting parallel wires separated by a distance $d$.

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$A$ bicycle wheel of radius $R$ has $n$ spokes. It is rotating at the rate of $F$ r.p.m. perpendicular to the horizontal component of earth's magnetic field $\vec{B}$. The e.m.f. induced between the rim and the centre of the wheel is

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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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$A$ thin wire of length $2 \ m$ is perpendicular to the $xy$-plane. It moves with a velocity $v = (2\hat{i} + 3\hat{j} + \hat{k}) \ m/s$ in a magnetic field $B = (\hat{i} + 2\hat{j}) \ Wb/m^2$. What is the induced potential difference (emf) across the ends of the wire?

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$A$ thin wire of length $2\,m$ is placed perpendicular to the $x-y$ plane. It is moved with velocity $\overrightarrow v = (2\hat i + 3\hat j + \hat k)\,m/s$ through a region of magnetic induction $\overrightarrow B = (\hat i + 2\hat j)\,Wb/m^2$. The potential difference induced between the ends of the wire is......$V$.

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Find the current in the wire for the configuration shown in the figure. Wire $PQ$ has negligible resistance. $\vec{B}$ is the magnetic field coming out of the paper. $\theta$ is a fixed angle made by $PQ$ as it travels smoothly over two conducting parallel wires separated by a distance $d$.

$A$ bicycle wheel of radius $R$ has $n$ spokes. It is rotating at the rate of $F$ r.p.m. perpendicular to the horizontal component of earth's magnetic field $\vec{B}$. The e.m.f. induced between the rim and the centre of the wheel is

$A$ thin wire of length $2 \ m$ is perpendicular to the $xy$-plane. It moves with a velocity $v = (2\hat{i} + 3\hat{j} + \hat{k}) \ m/s$ in a magnetic field $B = (\hat{i} + 2\hat{j}) \ Wb/m^2$. What is the induced potential difference (emf) across the ends of the wire?

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