There long straight wires A, B and C are carrying current as shown figure. Then the resultant force on B is directed 

(1) Towards A

(2) Towards C

(3) Perpendicular to the plane of paper and outward

(4) Perpendicular to the plane of paper and inward

Two long conductors, separated by a distance d carry current $I_1$ and $I_2$ in the same direction. They exert a force F on each other. Now the current in one of them is increased to two times and its direction is reversed. The distance is also increased to $3d$. The new value of the force between them is-

1. $-2F$                           

2.  $2F/3$

3. $-2F/3$                           

4. $-F/3$

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. $\frac{L{i}^{2}B}{4\mathrm{\pi }}$

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) $2n^{2}B$

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

Two identical long conducting wires AOB and COD are placed at right angle to each other, with one above other such that O is their common point for the two. The wires carry  I₁ and I₂ currents, respectively. Point P is lying at distance d from 0 along a direction perpendicular to the plane containing the wires. The magnetic field at the point P will be

(1) μ_o/2πd(I₁/I₂)

(2)μ_o/2πd (I₁+I₂)

(3)μ_o/2πd(I₁²-I₂²)

(4)μ_o/2πd(I₁²+I₂²)^1/2