Current i is carried in a wire of length L. If the wire is turned into a circular coil, the maximum magnitude of torque in a given magnetic field B will be:

1. $\frac{Li{B}^2}{2}$                     2. $\frac{L{i}^{2}B}{2}$

3. $\frac{L^{2}iB}{4\mathrm{\pi }}$                     4. $\frac{L{i}^{2}B}{4\mathrm{\pi }}$

Three long, straight parallel wires carrying current, are arranged as shown in figure. The force experienced by a 25 cm length of wire C is

1. ${10}^{-3}N$

2. $2.5\times {10}^{-3}N$

3. Zero

4. $1.5\times {10}^{-3}N$

An arrangement of three parallel straight wires placed perpendicular to the plane of paper carrying the same current I along the same direction as shown in the figure. Magnitude of force per unit length on the middle wire B is given by:

1. $\frac{{\mu }_o{i}^2}{2\mathrm{\pi d}}$

2. $\frac{2{\mu }_0{i}^2}{\mathrm{\pi d}}$

3. $\frac{\sqrt{2}{\mu }_o{i}^2}{\pi d}$

4. $\frac{{\mu }_o{i}^2}{\sqrt{2}\mathrm{\pi d}}$

A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic field at the center of loop is B. It is then bent into a circular coil of n turns. The magnetic field at the centre of this coil of n turns will be

1.  nB               
2.  $n^{2}B$
3.  2nB             
4. $2n^{2}B$

An electron is moving in a circular path under the influence of a transverse magnetic field of $3.57\times {10}^{-2}$ T. If the value of e/m is $1.76\times {10}^{11}$ C/kg, the frequency of revolution of the electron is

1. 1 GHz             

2. 100 MHz

3. 62.8 MHz         

4. 6.28 MHz

A square loop ABCD carrying a current i, is placed near and coplanar with a long straight conductor XY carrying a current I, the net force on the loop will be:

1. $\frac{{\mu }_{0}Ii}{2\mathrm{\pi }}$             
2. $\frac{2{\mu }_{0}IiL}{3\mathrm{\pi }}$

3. $\frac{{\mu }_{0}IiL}{2\mathrm{\pi }}$           
4. $\frac{2{\mu }_{0}Ii}{3\mathrm{\pi }}$

A wire carrying current l has the shape as shown in the adjoining figure. Linear parts of the wire are very long and parallel to X-axis while the semicircular portion of radius R is lying in the Y-Z plane. Magnetic field at point O is :


1. $\mathrm{B}=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\times \frac{\mathrm{i}}{\mathrm{R}}\left(\mathrm{\pi }\hat{\mathrm{i}}+2\hat{\mathrm{k}}\right)$

2. $\mathrm{B}=-\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\times \frac{\mathrm{i}}{\mathrm{R}}\left(\mathrm{\pi }\hat{\mathrm{i}}-2\hat{\mathrm{k}}\right)$

3. $\mathrm{B}=-\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\times \frac{\mathrm{i}}{\mathrm{R}}\left(\mathrm{\pi }\hat{\mathrm{i}}+2\hat{\mathrm{k}}\right)$

4. $\mathrm{B}=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\times \frac{\mathrm{i}}{\mathrm{R}}\left(\mathrm{\pi }\hat{\mathrm{i}}-2\hat{\mathrm{k}}\right)$

An electron moving in a circular orbit of radius r makes n rotations per second. The magnetic field produced at the centre has magnitude:

$1. \frac{{\mu }_{0}ne}{2\pi r}$
$2. Zero$
$3. \frac{n^{2}e}{r}$
$4. \frac{{\mu }_{0}ne}{2r}$

A proton and an alpha particle both enter a region of uniform magnetic field B, moving at right angles to the field B. If the radius of circular orbits for both the particles is equal and the kinetic energy acquired by proton is 1 MeV, the energy acquired by the alpha particle will be

(1)4 MeV

(2) 0.5 MeV

(3) 1.5 MeV

(4) 1 MeV