Two straight long conductors AOB and COD are perpendicular to each other and carry currents ${\mathrm{i}}_1$ and ${\mathrm{i}}_2$ . The magnitude of the magnetic induction at a point P at a distance a from the point O in a direction perpendicular to the plane ACBD is: 

1. $\frac{{\mathrm{\mu }}_0}{2\mathrm{\pi a}}({\mathrm{i}}_1+{\mathrm{i}}_2)$                    2. $\frac{{\mathrm{\mu }}_0}{2\mathrm{\pi a}}({\mathrm{i}}_1-{\mathrm{i}}_2)$

3. $\frac{{\mathrm{\mu }}_0}{2\mathrm{\pi a}}{\left({\mathrm{i}}_1^2+{\mathrm{i}}_2^2\right)}^{1/2}$                4. $\frac{{\mathrm{\mu }}_0}{2\mathrm{\pi a}}\frac{{\mathrm{i}}_1{\mathrm{i}}_2}{({\mathrm{i}}_1+{\mathrm{i}}_2)}$

In the given figure, the electron enters into the magnetic field. It deflects in ...... direction

1. + ve X direction

2. – ve X direction

3. + ve Y direction

4. – ve Y direction

A cell is connected between the points A and C of a circular conductor ABCD of centre O with angle AOC = $60°$ . If ${\mathrm{B}}_1$ and ${\mathrm{B}}_2$ are the magnitudes of the magnetic fields at O due to the currents in ABC and ADC respectively, the ratio $\frac{{\mathrm{B}}_1}{{\mathrm{B}}_2}$ is:

1. 0.2                             

2. 6                                    

3. 1

4. 5                    

An electron, a proton, a deuteron and an alpha particle, each having the same speed are in a region of constant magnetic field perpendicular to the direction of the velocities of the particles. The radius of the circular orbits of these particles are respectively $R_e$, $R_p$, $R_d$ and $R_{\alpha }$. It follows that 

1. $R_e=R_p$               

2. $R_p=R_d$

3. $R_d=R_{\alpha }$                 

4. $R_p=R_{\alpha }$

An infinitely long conductor PQR is bent to form a right angle as shown. A current I flows through PQR. The magnetic field due to this current at the point M is H₁. Now another infinitely long straight conductor QS is connected at Q so that the current is I/2 in QR as well as in QS, The current in PQ remaining unchanged. The magnetic field at M is now $H_{\mathit{2}}$The ratio $H_{\mathit{1}}\mathit{/}{H}_{\mathit{2}}$ is given by

1. $\frac{1}{2}$                                                

2.  1

3.  $\frac{2}{3}$

4.  2

A very long straight wire carries a current I. At the instant when a charge +Qat point P has velocity $\overrightarrow{v}$ , as shown, the force on the charge is:

1. Opposite to OX                            
2. Along OX
3. Opposite to OY                             
4. Along OY

An electric field of 1500 V / m and a magnetic field of 0.40 weber / $mete{r}^2$ act on a moving electron. The minimum uniform speed along a straight line the electron could have is

1. $1.6\times {10}^{15}m/s$                         

2. $6\times {10}^{-16}m/s$

3. $3.75\times {10}^{3}m/s$                         

4. $3.75\times {10}^{2}m/s$

A coil having N turns is wound tightly in the form of a spiral with inner and outer radii a and b respectively. When a current I passes through the coil, the magnetic field at the centre is:

1. $\frac{{\mathrm{\mu }}_0\mathrm{NI}}{\mathrm{b}}$                                  2. $\frac{2{\mathrm{\mu }}_0\mathrm{NI}}{\mathrm{a}}$

3. $\frac{{\mathrm{\mu }}_0\mathrm{NI}}{2(\mathrm{b}-\mathrm{a})}\ln \frac{\mathrm{b}}{\mathrm{a}}$                         4. $\frac{{\mathrm{\mu }}_0{\mathrm{I}}^{\mathrm{N}}}{2(\mathrm{b}-\mathrm{a})}\ln \frac{\mathrm{b}}{\mathrm{a}}$

An electron, moving in a uniform magnetic field of induction of intensity $\stackrel{\rightharpoonup }{B,}$ has its radius directly proportional to :

1. Its charge                       

2. Magnetic field

3. Speed                               

4. None of these