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 = ${\mathrm{\mu }}_0$)

Two particles X and Y having equal charges, after being accelerated through the same potential difference, enter a region of uniform magnetic field and describes circular path of radius ${\mathrm{R}}_1$ and ${\mathrm{R}}_2$ respectively. The ratio of mass of X to that of Y is :

(a)  ${\left(\frac{{\mathrm{R}}_1}{{\mathrm{R}}_2}\right)}^{1/2}$                         (b)  $\frac{{\mathrm{R}}_2}{{\mathrm{R}}_1}$

(c)  ${\left(\frac{{\mathrm{R}}_1}{{\mathrm{R}}_2}\right)}^2$                             (d)  $\frac{{\mathrm{R}}_1}{{\mathrm{R}}_2}$  

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

If the direction of the initial velocity of the charged particle is perpendicular to the magnetic field, then the orbit will be
                                                                      or
The path executed by a charged particle whose motion is perpendicular to magnetic field is :

(1) A straight line                               

(2) An ellipse

(3) A circle                                         

(4) A helix

A proton and an α- particle enter a uniform magnetic field perpendicularly at the same speed. If a proton takes 25 μs to make 5 revolutions, then the periodic time for the α-particle will be:

1. 50 $\mathrm{\mu }$s               

2. 25 $\mathrm{\mu }$s 

3. 10 $\mathrm{\mu }$s                 

4. 5 $\mathrm{\mu }$s

An $\mathrm{\alpha }-$ particle travels in a circular path of radius 0.45 m in a magnetic field $\mathrm{B}=1.2 Wb\mathit{/}{m}^{\mathit{2}}$ with a speed of $2.6\times {10}^7 m\mathit{/}sec$ . The period of revolution of the $\mathrm{\alpha }-$ particle is :

(a)  $1.1\times {10}^{-5 }$ sec          (b)  $1.1\times {10}^{-6}$ sec

(c)  $1.1\times {10}^{-7}$ sec           (d)  $1.1\times {10}^{-8}$ sec

A rectangular loop carrying a current i is situated near a long straight wire such that the wire is parallel to the one of the sides of the loop and is in the plane of the loop. If a steady current I is established in wire as shown in figure, the loop will

(1) Rotate about an axis parallel to the wire

(2) Move away from the wire or towards right

(3) Move towards the wire

(4) Remain stationary