An electron and a proton with equal momentum enter perpendicularly into a uniform magnetic field, then :

(1) The path of proton shall be more curved than that of electron

(2) The path of proton shall be less curved than that of electron

(3) Both are equally curved

(4) Path of both will be a straight line

One proton beam enters a magnetic field of ${10}^{-4}T$ normally, Specific charge = ${10}^{11}C/kg.$ velocity = ${10}^{7}m/s.$ What is the radius of the circle described by it:

1. $0.1m$                               

2. $1m$

3. $10m$                               

4. None of these

In a cyclotron, the angular frequency of a charged particle is independent of :

1. Mass                         

2.  Speed

3. Charge                       

4. Magnetic field

The maximum kinetic energy of the positive ion of charge q and mass m in the cyclotron of radius \(r_o\) in which applied magnetic field is B, is:

1. $\frac{q^{2}B{r}_0}{2m}$                        

2. $\frac{q{B}^2{r}_0}{2m}$

3. $\frac{q^2{B}^2{r}_0^2}{2m}$                        

4. $\frac{qB{r}_0}{2m^2}$

An electron of mass m and charge q is traveling with a speed v along a circular path of radius r at right angles to a uniform of the magnetic field B. If the speed of the electron is doubled and the magnetic field is halved, then resulting path would have a radius of:

1. $\frac{r}{4}$                           

2. $\frac{r}{2}$ 

3. $2r$                           

4. $4r$

Cyclotron cannot be used to accelerate

(1) Electrons                       

(2) Neutrons

(3) Positive ions                   

(4) Both (1) and (2)

Two particles A and B of masses $m_A$ and $m_B$ respectively and having the same type of charge are moving in a plane. A uniform magnetic field exists perpendicular to this plane. The speeds of the particles are $v_A$ and $v_B$ respectively, and the trajectories are as shown in the figure. Then

(1) $m_A{v}_A<m_B{v}_B$                              

(2) $m_A{v}_A>m_B{v}_B$

(3) $m_A<m_B and v_A<v_B$                  

(4) $m_A=m_B and v_A=v_B$

Magnetic field due to a ring having n turns at a distance x on its axis is proportional to (if r = radius of ring) :

(1)  $\frac{\mathrm{r}}{({\mathrm{x}}^2+{\mathrm{r}}^2)}$                                     

(2)  $\frac{{\mathrm{r}}^2}{({\mathrm{x}}^2+{\mathrm{r}}^2{)}^{3/2}}$

(3)  $\frac{{\mathrm{nr}}^2}{({\mathrm{x}}^2+{\mathrm{r}}^2{)}^{3/2}}$                                 

(4)  $\frac{{\mathrm{n}}^2{\mathrm{r}}^2}{({\mathrm{x}}^2+{\mathrm{r}}^2{)}^{3/2^{}}}$ 

A and B are two concentric circular conductors of centre O and carrying currents ${\mathrm{i}}_{1 }$and ${\mathrm{i}}_2$ as shown in the adjacent figure. If ratio of their radii is 1 : 2 and ratio of the flux densities at O due to A and B is 1 : 3, then the value of ${\mathrm{i}}_1/{\mathrm{i}}_2$ is

(a)  $\frac{1}{6}$                       (b)  $\frac{1}{4}$                                

 (c)  $\frac{1}{3}$                       (d)  $\frac{1}{2}$                                                                                                          

PQRS is a square loop made of uniform conducting wire the current enters the loop at P and leaves at S. Then the magnetic field will be

(a) Maximum at the centre of the loop
(b) Zero at the centre of the loop
(c) Zero at all points inside the loop
(d) Zero at all points outside of the loop