Q. No.A metal rod of length \(L\) \(\omega\). \(B(r)=B_{0} {e}^{-\lambda r} ; \lambda\)
1. \(B_0 \omega\left[\dfrac{1}{\lambda^2}-e^{-\lambda L}\left(\dfrac{1}{\lambda^2}+\dfrac{L}{\lambda}\right)\right]\)
2. \(B_0 \omega\left[\dfrac{1}{\lambda^2}+e^{-\lambda L}\left(\dfrac{1}{\lambda^2}+\dfrac{L}{\lambda}\right)\right]\)
3. \(B_0 \omega\left[\dfrac{4}{\lambda^2}-e^{-2 \lambda L}\left(\dfrac{1}{\lambda^2}+\dfrac{2 L}{\lambda}\right)\right]\)
4. \(B_0 \omega\left[\dfrac{3}{\lambda^2}-e^{-3 \lambda L}\left(\dfrac{3}{\lambda^2}+\dfrac{L}{\lambda}\right)\right]\)
1. \(B_0 \omega\left[\dfrac{1}{\lambda^2}-e^{-\lambda L}\left(\dfrac{1}{\lambda^2}+\dfrac{L}{\lambda}\right)\right]\)
2. \(B_0 \omega\left[\dfrac{1}{\lambda^2}+e^{-\lambda L}\left(\dfrac{1}{\lambda^2}+\dfrac{L}{\lambda}\right)\right]\)
3. \(B_0 \omega\left[\dfrac{4}{\lambda^2}-e^{-2 \lambda L}\left(\dfrac{1}{\lambda^2}+\dfrac{2 L}{\lambda}\right)\right]\)
4. \(B_0 \omega\left[\dfrac{3}{\lambda^2}-e^{-3 \lambda L}\left(\dfrac{3}{\lambda^2}+\dfrac{L}{\lambda}\right)\right]\)
Subtopic: Motional emf |
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A \(1~\text{m}\) long metal rod \(AB\) completes the circuit as shown in figure. The area of circuit is perpendicular to the magnetic field of \(0.10~\text{T}\). If the resistance of the total circuit is \(2~\Omega\), then the force needed to move the rod towards right with constant speed \((v)\) of \(1.5~\text{m/s}\) is: (in N)
1. \(7.5 \times 10^{-2}\)
2. \(5.7 \times 10^{-3}\)
3. \(5.7 \times 10^{-2}\)
4. \(7.5 \times 10^{-3}\)
1. \(7.5 \times 10^{-2}\)
2. \(5.7 \times 10^{-3}\)
3. \(5.7 \times 10^{-2}\)
4. \(7.5 \times 10^{-3}\)
Subtopic: Motional emf |
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\(XPQY\) is a vertical smooth long loop having a total resistance \(R\) where \(PX\) is parallel to \(QY\) and separation between them is \(l\). A constant magnetic field \(B\) perpendicular to the plane of the loop exists in the entire space. A rod \(CD\) of length \(L(L>l)\) and mass \(m\) is made to slide down from rest under gravity as shown in the figure. The terminal speed (in m/s) acquired by the rod is: (\(g\) = acceleration due to gravity)
1. \(\dfrac{2 {mgR}}{{B}^2 l^2}\)
2. \(\dfrac{8 {mgR}}{{B}^2 l^2}\)
3. \(\dfrac{2 {mgR}}{{B}^2 {L}^2}\)
4. \(\dfrac{{mgR}}{{B}^2 l^2}\)
1. \(\dfrac{2 {mgR}}{{B}^2 l^2}\)
2. \(\dfrac{8 {mgR}}{{B}^2 l^2}\)
3. \(\dfrac{2 {mgR}}{{B}^2 {L}^2}\)
4. \(\dfrac{{mgR}}{{B}^2 l^2}\)
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A conducting circular loop is rotated about its diameter at a constant angular speed of \(100~\text{rad/s}\) in a magnetic field of \(0.5~\text{T}\) perpendicular to the axis of rotation. When the loop is rotated by \(30^\circ\) from the horizontal position, the induced EMF is \(15.4~\text{mV}.\) The radius of the loop is: (in mm)
\(\left(\text {Take } \pi=\frac{22}{7}\right)\)
1. \(10\)
2. \(14\)
3. \(19\)
4. \(20\)
\(\left(\text {Take } \pi=\frac{22}{7}\right)\)
1. \(10\)
2. \(14\)
3. \(19\)
4. \(20\)
Subtopic: Motional emf |
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A \(20\) m long uniform copper wire held horizontally is allowed to fall under the gravity (\(g=10\) m/s2) through a uniform horizontal magnetic field of \(0.5\) Gauss perpendicular to the length of the wire. The induced EMF across the wire it travels a vertical distance of \(200\) m is: (in mV)
1. \(0.2 \sqrt{10}\)
2. \(20 \sqrt{10}\)
3. \(2 \sqrt{10}\)
4. \(200 \sqrt{10}\)
1. \(0.2 \sqrt{10}\)
2. \(20 \sqrt{10}\)
3. \(2 \sqrt{10}\)
4. \(200 \sqrt{10}\)
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A simple pendulum made of mass \(10~\text{g}\) and a metallic wire of length \(10~\text{cm}\) is suspended vertically in a uniform magnetic field of \(2~\text{T}\). The magnetic field direction is perpendicular to the plane of oscillations of the pendulum. If the pendulum is released from an angle of \(60^{\circ}\) with vertical, then maximum induced EMF between the point of suspension and point of oscillation is: (in mV) (Take \(g=10\) m/s²)
1. \(100\)
2. \(240\)
3. \(300\)
4. \(350\)
1. \(100\)
2. \(240\)
3. \(300\)
4. \(350\)
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Conductor wire \(ABCDE \) with each arm \(10 ~\text{cm}\) in length is placed in magnetic field of \(1/\sqrt{2}~\text{Tesla}\) , perpendicular to its plane. When conductor is pulled towards right with constant velocity of \(10 ~\text{cm/s,}\) induced emf between points \(A\) and \(E \) is:
1. \(39~\text{mV}\)
2. \(10~\text{mV}\)
3. \(36~\text{mV}\)
4. \(27~\text{mV}\)
1. \(39~\text{mV}\)
2. \(10~\text{mV}\)
3. \(36~\text{mV}\)
4. \(27~\text{mV}\)
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A conducting bar moves on two conducting rails as shown in the figure. A constant magnetic field \(B\) exists into the page. The bar starts to move from the vertex at time \(t=0 \) with a constant velocity. If the induced \(EMF\) is \(E \propto t^n,\) then value of \(n\) is:
1. \(1\)
2. \(2\)
3. \(3\)
4. \(4\)
1. \(1\)
2. \(2\)
3. \(3\)
4. \(4\)
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A circular copper disc of radius \(20~\text{cm}\) rotates with a uniform angular velocity of \(10 \pi ~\text{rad s}^{-1}\) about an axis passing through its centre and perpendicular to its plane. The disc is placed in a uniform magnetic field of \(0.4~ \text T,\) directed perpendicular to its plane and parallel to the axis of rotation. The magnitude of the potential difference developed between the centre and the rim of the disc is: \((\text{use}~\pi=3.14)\)
1. \(0.2512~\text V \)
2. \( 0.0628~\text{V} \)
3. \(0.1256 ~\text V \)
4. \( 0.5024 ~\text V\)
Subtopic: Motional emf |
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A coil of area \(A\) and \(N\) turns is rotating with angular velocity \(ω\) in a uniform magnetic field \(\vec{B}\) about an axis perpendicular to \(\vec{B}.\) Magnetic flux \(\phi\) and induced emf \(\varepsilon\) across it, at an instant when
1. \(\phi={AB}, \varepsilon ={NAB} \omega~\)
2. \(\phi=0, \varepsilon={NAB} \omega~\)
3. \(\phi={AB}, \varepsilon=0~\)
4. \(\phi=0, \varepsilon=0~\)
1. \(\phi={AB}, \varepsilon ={NAB} \omega~\)
2. \(\phi=0, \varepsilon={NAB} \omega~\)
3. \(\phi={AB}, \varepsilon=0~\)
4. \(\phi=0, \varepsilon=0~\)
Subtopic: Motional emf |
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