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. \(\dfrac{{mgR}}{{B}^2 l^2}\)
Subtopic: Motional emf |
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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. \(200 \sqrt{10}\)
Subtopic: Motional emf |
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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. \(350\)
Subtopic: Motional emf |
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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. \(4\)
Subtopic: Motional emf |
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A uniform magnetic field of \(0.4~ \text T~\) acts perpendicular to a circular copper disc \(20~\text{cm}\) in radius. The disc is having a uniform angular velocity of \(10 \pi ~\text{rad s}^{-1}\) about an axis through its centre and perpendicular to the disc. What is the potential difference developed between the axis of the disc and the rim? \((\text{use}~\pi=3.14)\)
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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