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

An electric current passes through a long straight wire. At a distance 5 cm from the wire, the magnetic field is B. The field at 20 cm from the wire would be :

(1) $\frac{\mathrm{B}}{6}$                   

(2) $\frac{\mathrm{B}}{4}$

(3) $\frac{\mathrm{B}}{3}$                     

(4) $\frac{\mathrm{B}}{2}$ 

 The dimension of the magnetic field intensity B is:

1.  ${\mathrm{MLT}}^{-2}{\mathrm{A}}^{-1}$                   

2.  ${\mathrm{MT}}^{-2}{\mathrm{A}}^{-1}$

3.  ${\mathrm{ML}}^2{\mathrm{TA}}^{-2}$                       

4.  ${\mathrm{M}}^2{\mathrm{LT}}^{-2}{\mathrm{A}}^{-1}$

If a wire in the form of a square with a side ‘a’ carries a current i, then the magnetic induction at the centre of the square wire will be:
(Magnetic permeability of free space = ${\mathrm{\mu }}_0$)

A beam of ions with velocity $2\times {10}^5\mathit{ }m\mathit{/}s$ enters normally into a uniform magnetic field of $4\times {10}^{-2}$tesla. If the specific charge of the ion is $5\times {10}^7$ C/kg , then the radius of the circular path described will be :

(a) 0.10 m              (b)  0.16 m

(c)  0.20 m              (d) 0.25 m