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Q. No.A long thin wire of length \(L\) carrying a current \(i,\) is wrapped uniformly around a long solenoid of volume \(V.\) The radius of the solenoid is \(r.\) The magnetic field at its centre is:
1.
\(\dfrac{\mu_0~L~r~i}{V}\)
2.
\(\dfrac{\mu_0~L~r~i}{2\pi~V}\)
3.
\(\dfrac{\mu_0~L~r~i}{2V}\)
4.
\(\dfrac{\mu_0~L~r~i}{4\pi~V}\)
Subtopic: Magnetic Field due to various cases |
51%
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A long cylindrical solenoid of length \(L\) and total number of turns \(N\) produces a magnetic field \(B_0\) at its centre with a current of \(1\) A flowing through its coils. If the same current of \(1\) A is sent through a circular wire of radius \(R\) then the same field \(B_0\) is produced at its centre. Then, \(\dfrac{R}{L}\) equals:
| 1. | \(N\) | 2. | \(\dfrac 1N\) |
| 3. | \(2N\) | 4. | \(\dfrac{1}{2N}\) |
Subtopic: Magnetic Field due to various cases |
74%
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A charged particle moves in a circular path of radius \(r\) in a uniform magnetic field \(B,\) perpendicular to the plane of motion. The same particle is observed to move in a circular path around an infinite line charge \(\lambda\) (charge/unit length), moving with the same kinetic energy as before. The charge to mass ratio of the particle is proportional to:
| 1. | \(\lambda Br\) | 2. | \(\dfrac{\lambda Br}{r}\) |
| 3. | \(\dfrac{\lambda}{Br}\) | 4. | \(\dfrac{\lambda}{B^2r^2}\) |
Subtopic: Lorentz Force |
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A charged particle of charge \(q,\) mass \(m\) moves in a circular path under the action of a uniform magnetic field, whose flux through this path is \(\phi.\) The magnetic moment due to the particle's motion is:
| 1. | \(\dfrac{q^2\phi}{2m}\) | 2. | \(\dfrac{q^2\phi}{2\pi m}\) |
| 3. | \(\dfrac{q^2\phi}{m}\) | 4. | \(\dfrac{q^2\phi}{\pi m}\) |
Subtopic: Magnetic Moment |
71%
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A current-carrying wire is placed in a uniform magnetic field and the force on the wire is measured at different angular positions of the wire, as it is rotated in the \(x-y\) plane. Initially, the wire is along the \(x\)-axis. The magnitude of the magnetic force\((F)\) is plotted as a function of the angle\((\theta)\) made by the current-carrying wire with the \(x\)-axis.
Which of the following is the possible magnetic field (in tesla)?
1. \(2\hat j\)
2. \(2\hat k\)
3. \(2\hat i+2\hat k\)
4. \(2\hat j+2\hat k\)
Which of the following is the possible magnetic field (in tesla)?
1. \(2\hat j\)
2. \(2\hat k\)
3. \(2\hat i+2\hat k\)
4. \(2\hat j+2\hat k\)
Subtopic: Magnetic Field due to various cases |
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A long straight current-carrying wire is placed along the axis of a solenoid and it is found that the field within the solenoid at a distance of \(\dfrac r{10}\) from the wire is doubled when a current \(I\) passes through the wire and the solenoid, \(r\) being the radius of the solenoid. The number of turns per unit length of the solenoid are:
| 1. | \(\dfrac{5}{\sqrt3r}\) | 2. | \(\dfrac{5}{\sqrt3\pi r}\) |
| 3. | \(\dfrac{5}{r}\) | 4. | \(\dfrac{5}{\pi r}\) |
Subtopic: Magnetic Field due to various cases |
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An ammeter having a resistance of \(0.1~ \Omega\) can measure a maximum current of \(2~\text{A}\). To convert it into a voltmeter that can measure upto \(50~\text{V},\) one must add:
| 1. | a resistance of \(25~\Omega\) in series |
| 2. | a resistance of \(\dfrac1{25}~\Omega\) in series |
| 3. | a resistance of \(25~\Omega\) in parallel |
| 4. | a resistance of \(\dfrac1{25}~\Omega\) in parallel |
Subtopic: Conversion to Ammeter & Voltmeter |
74%
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A charged particle (charge: \(q\)) moves in a circular orbit in a uniform magnetic field, its orbit enclosing a magnetic flux \(\Phi.\) The angular momentum of the particle is:
| 1. | \(q\Phi\) | 2. | \(\dfrac{q\Phi}{2\pi}\) |
| 3. | \(\pi q\Phi\) | 4. | \(\dfrac{q\Phi}{\pi}\) |
Subtopic: Lorentz Force |
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A galvanometer \(G\) (having very small resistance), when connected with a resistance of \(10~\text k\Omega\) in series, can function as a voltmeter measuring a maximum voltage of \(20\) V. The current required to give a full scale deflection on the galvanometer is:
1. \(0.1\) mA
2. \(0.2\) mA
3. \(1\) mA
4. \(2\) mA
1. \(0.1\) mA
2. \(0.2\) mA
3. \(1\) mA
4. \(2\) mA
Subtopic: Moving Coil Galvanometer |
79%
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A conducting wire is bent into the form of a square \(ABCD,\) and electrical connections are established in two different ways:
The same potential difference is established between the two connected ends. Current is, however, allowed to take only a single path from the positive to the negative terminal by disconnecting the other path. Let the magnetic field at the centre in these cases be \(B_\text I,B_\text{II}.\) Then, \(\frac{B_\text I}{B_\text{II}}=\)
| (I) | at two adjacent vertices \(A,B\) |
| (II) | at two points \(A,C\) at the ends of a diagonal. |
| 1. | \(2\) | 2. | \(\dfrac12\) |
| 3. | \(\dfrac{1}{\sqrt2}\) | 4. | \(1\) |
Subtopic: Magnetic Field due to various cases |
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