Two identically charged particles A and B initially at rest, are accelerated by a common potential difference V. They enter into a transverse uniform magnetic field B. If they describe a circular path of radii  $r_1$ $and$ $r_2$ respectively, then their mass ratio is:

1.  ${\left(\frac{r_1}{r_2}\right)}^2$

2. ${\left(\frac{r_2}{r_1}\right)}^2$

3. $\left(\frac{r_1}{r_2}\right)$

4. $\left(\frac{r_2}{r_1}\right)$

A charge having q/m equal to 10⁸ c/kg and with velocity 3 × 10⁵ m/s enters into a uniform magnetic field B = 0.3 tesla at an angle 30º with the direction of field. Then the radius of curvature will be:

1. 0.01 cm

2. 0.5 cm

3. 1 cm

4. 2 cm

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. $\frac{2m}{eBE}$

In the Thomson mass spectrograph where \(\vec{E}\perp\vec{B}\) 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. \(\frac{E^{2}}{B^{2}}\)

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

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

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

4. \(q\vec{B}+{q}(\vec{E} \times \vec{v})\)

A very long straight wire carries a current I. At the instant when a charge +Q at point P has velocity $\overrightarrow{\mathrm{v}}$, as shown, the force on the charge is

1. Along ox

2. Opposite to oy

3. Along oy

4. Opposite to ox