A charged particle moves through a magnetic field in a direction perpendicular to it. Then:

1. the speed of the particle remains unchanged.

2. the direction of the particle remains unchanged.

3. the acceleration remains unchanged.

4. the velocity remains unchanged.

A long solenoid carrying a current produces a magnetic field \(B\) along its axis. If the current is doubled and the number of turns per cm is halved, what will be the new value of the magnetic field?
1. \(B/2\)
2. \(B\)
3. \(2B\)
4. \(4B\)

If the number of turns, area, and current through a coil are given by \(n\), \(A\) and \(i\) respectively then its magnetic moment will be: 
1. \(niA\)
2. \(n^{2}iA\)
3. \(niA^{2}\)
4. \(\frac{ni}{\sqrt{A}}\)

The tangent galvanometer is used to measure: 
1.  Potential difference
2.  Current
3.  Resistance
4.  In measuring the charge

In the Thomson mass spectrograph where $\overrightarrow{E}\perp \overrightarrow{B}$ the velocity of the undeflected electron beam will be:

1.  \(\frac{\left| \vec{E}\right|}{\left|\vec{B} \right|}\)
2. \(\vec{E}\times \vec{B}\)
3. \(\frac{\left| \vec{B}\right|}{\left|\vec{E} \right|}\)
4. \(\frac{E^{2}}{B^{2}}\)

An electron having mass 'm' and kinetic energy E enter in a uniform magnetic field B perpendicularly. Its frequency will be:

1. $\frac{eE}{qVB}$

2. $\frac{2\mathrm{\pi m}}{eB}$

3. $\frac{eB}{2\mathrm{\pi m}}$

4. $\frac{2m}{eBE}$

If a charge '\(q\)' moves with velocity \(\mathrm{v}\), in a region where electric field (\(\mathrm{E}\)) and magnetic field (\(\mathrm{B}\)) both exist, then force on it is: 
1. \(q(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})\)

2. \(q \overrightarrow{\mathrm{E}}+\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})\)

3. \( q \overrightarrow{\mathrm{E}}+q(\overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{v}})\)

4. \(\mathrm{qB}+\mathrm{q}(\overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{v}})\)

The magnetic field of a given length of wire for a single-turn coil at its centre is 'B'. Its value for two turns coil for the same wire will be:

1. $\frac{\mathrm{B}}{4}$

2. $\frac{\mathrm{B}}{2}$

3. $4\mathrm{B}$

4. $2\mathrm{B}$

To convert a galvanometer into a voltmeter one should connect a:

A thick current-carrying cable of radius '\(R\)' carries current \('I'\) uniformly distributed across its cross-section. The variation of magnetic field \(B(r)\) due to the cable with the distance '\(r\)' from the axis of the cable is represented by:

1.		2.
3.		4.