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. \(2\hat j+2\hat k\)

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 |
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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
Subtopic:  Moving Coil Galvanometer |
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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:
(I) at two adjacent vertices \(A,B\)
(II) at two points \(A,C\) at the ends of a diagonal.
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, \(\frac{B_\text I}{B_\text{II}}=\)
1. \(2\) 2. \(\dfrac12\)
3. \(\dfrac{1}{\sqrt2}\) 4. \(1\)
Subtopic:  Magnetic Field due to various cases |
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A charged particle of charge \(q,\) mass \(m\) moves in a circular path in a uniform magnetic field \(B.\) The current due to the motion of this particle has the average value:
1. \(\dfrac{q^2B}{2\pi m}\) 2. \(\dfrac{qB^2}{m}\)
3. \(\dfrac{2\pi m}{q^2B}\) 4. \(\dfrac{qB}{m^2}\)
Subtopic:  Lorentz Force |
 79%
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Two long parallel wires carrying currents \(I_1\) and \(I_2\) give a magnetic field of \(3\) G at a point exactly mid-way between the two wires. When one of the currents is reversed, the field becomes \(5\) G. The ratio of the large current to the smaller one is:
1. \(2\) 2. \(\dfrac43\)
3. \(\dfrac32\) 4. \(4\)
Subtopic:  Magnetic Field due to various cases |
 75%
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Given below are two statements: 
Statement I: A charged particle moving in a magnetic field experiences a force which is zero only when it moves in the direction of the field or against it.
Statement II: Whenever a charged particle moves in a uniform magnetic field, its trajectory may be a circle, a straight line or a helix.
 
1. Statement I is incorrect and Statement II is correct.
2. Both Statement I and Statement II are correct.
3. Both Statement I and Statement II are incorrect.
4. Statement I is correct and Statement II is incorrect.
Subtopic:  Lorentz Force |
 73%
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A current \(I,\) flowing through a long straight wire produces a magnetic field of \(4\times10^{-3}\) T at a point \(P\)​​​​, \(20\) cm away. If the wire is replaced by a circular wire of radius \(20\) cm, centred at \(P\) and carrying the same current \(I,\) then the magnetic field at \(P\) will be:
1. \(4\pi\times10^{-3}\) T
2.  \(\dfrac4\pi\times10^{-3}\) T
3.  \(4\times10^{-3} \) T
4.  \(2\times10^{-3}\) T
Subtopic:  Magnetic Field due to various cases |
 70%
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