Q. No.A bar magnet is released along the vertical axis of the conducting coil. The acceleration of the bar magnet is:
1.
greater than \(g\).
2.
less than \(g\).
3.
equal to \(g\).
4.
zero.
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A rod having length \(l\) and resistance \(R_0\) is moving with a speed \(v\) as shown in the figure. The current through the rod is:
1. \(\dfrac{B l v}{\frac{R_{1} R_{2}}{R_{1} + R_{2}} + R_{0}}\)
2. \(\dfrac{Blv}{\left(\frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{o}}\right)^{2}}\)
3. \(\dfrac{B l v}{R_{1} + R_{2} + R_{0}}\)
4. \(\dfrac{B l v}{\frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{0}}}\)
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1. \(\dfrac{E^{2}}{2 R}\)
2. \(\dfrac{E^{2} L}{2 R^{2}}\)
3. \(\dfrac{E^{2} L}{R}\)
4. \(\dfrac{E^{2} L}{2 R}\)
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The coefficient of mutual inductance between two coils depends upon:
| 1. | medium between coils |
| 2. | separation between coils |
| 3. | orientation of coils |
| 4. | All of these |
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| 1. | \(\dfrac{L}{l}\) | 2. | \(\dfrac{l}{L}\) |
| 3. | \(\dfrac{L^2}{l}\) | 4. | \(\dfrac{l^2}{L}\) |
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Two coils have a mutual inductance of \(5\) mH. The current changes in the first coil according to the equation \(I=I_{0}\cos\omega t,\) where \(I_{0}=10~\text{A}\) and \(\omega = 100\pi ~\text{rad/s}\). The maximum value of emf induced in the second coil is:
1. \(5\pi~\text{V}\)
2. \(2\pi~\text{V}\)
3. \(4\pi~\text{V}\)
4. \(\pi~\text{V}\)
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Eddy currents are used in:
1. Induction furnace
2. Electromagnetic brakes
3. Speedometers
4. All of these
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The magnetic flux linked with a coil varies with time as \(\phi = 2t^2-6t+5,\) where \(\phi \) is in Weber and \(t\) is in seconds. The induced current is zero at:
| 1. | \(t=0\) | 2. | \(t= 1.5~\text{s}\) |
| 3. | \(t=3~\text{s}\) | 4. | \(t=5~\text{s}\) |
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If a current is passed through a circular loop of radius \(R\) then magnetic flux through a coplanar square loop of side \(l\) as shown in the figure \((l<<R)\) is:
1. \(\dfrac{\mu_{0} I}{2} \dfrac{R^{2}}{l}\)
2. \(\dfrac{\mu_{0} I l^{2}}{2 R}\)
3. \(\dfrac{\mu_{0}I \pi R^{2}}{2 l}\)
4. \(\dfrac{\mu_{0} \pi R^{2} I}{l}\)
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The radius of a loop as shown in the figure is \(10~\text{cm}.\) If the magnetic field is uniform and has a value \(10^{-2}~ \text{T},\) then the flux through the loop will be:
1. \(2 \pi \times 10^{-2}~\text{Wb}\)
2. \(3 \pi \times 10^{-4}~\text{Wb}\)
3. \(5 \pi \times 10^{-5}~\text{Wb}\)
4. \(5 \pi \times 10^{-4}~\text{Wb}\)
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