A current of 5 amperes is flowing in a wire of length 1.5 meters. A force of 7.5 N acts on it when it is placed in a uniform magnetic field of 2 Tesla. The angle between the magnetic field and the direction of the current is:

1. 30$°$                     2. 45°

3. 60°                     4. 90°

The field normal to the plane of a coil of n turns and radius r which carries a current i is measured on the axis of the coil at a small distance h from the centre of the coil. This is smaller than the field at the centre by the fraction

(1) $\frac{3}{2}\frac{{\mathrm{h}}^2}{{\mathrm{r}}^2}$                       

(2) $\frac{2}{3}\frac{{\mathrm{h}}^2}{{\mathrm{r}}^2}$

(3)  $\frac{3}{2}\frac{{\mathrm{r}}^2}{{\mathrm{h}}^2}$                       

(4) $\frac{2}{3}\frac{{\mathrm{r}}^2}{{\mathrm{h}}^2}$ 

A proton of mass $1.67\times {10}^{-27}kg$ and charge $1.6\times {10}^{-19}C$ is projected with a speed of $2\times {10}^{6}m/s$ at an angle of $60°$ to the X-axis. If a uniform magnetic field of 0.104 Tesla is applied along Y-axis, the path of the proton is:

1. A circle of radius = 0.2 m and time period $\mathrm{\pi }\times {10}^{-7}\mathrm{s}$

2. A circle of radius = 0.1 m and time period $2\mathrm{\pi }\times {10}^{-7}\mathrm{s}$ 

3. A helix of radius = 0.1 m and time period $2\mathrm{\pi }\times {10}^{-7}\mathrm{s}$

4. A helix of radius = 0.2 m and time period $4\mathrm{\pi }\times {10}^{-7}\mathrm{s}$

The magnetic field at the centre of a circular coil of radius r is $\mathrm{\pi }$ times that due to a long straight wire at a distance r from it, for equal currents. Figure here shows three cases : in all cases the circular part has radius r and straight ones are infinitely long. For same current the B field at the centre P in cases 1, 2, 3 have the ratio

(a) $\left(\frac{-\mathrm{\pi }}{2}\right)\mathbf{:}\left(\frac{\mathrm{\pi }}{2}\right)\mathbf{ }\mathbf{:}\left(\frac{3\mathrm{\pi }}{4}-\frac{1}{2}\right)$

(b) $\left(-\frac{\mathrm{\pi }}{2}+1\right)\mathbf{:}\left(\frac{\mathrm{\pi }}{2}+1\right)\mathbf{:}\left(\frac{3\mathrm{\pi }}{4}+\frac{1}{2}\right)$

(c) $-\frac{\mathrm{\pi }}{2}\mathbf{:}\frac{\mathrm{\pi }}{2}\mathbf{:}3\frac{\mathrm{\pi }}{4}$

(d) $\left(-\frac{\mathrm{\pi }}{2}-1\right)\mathbf{:}\left(\frac{\mathrm{\pi }}{2}-\frac{1}{4}\right)\mathbf{:}\left(\frac{3\mathrm{\pi }}{4}+\frac{1}{2}\right)$

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

(a) $\frac{1}{2}$                                                

(b)  1

(c)  $\frac{2}{3}$

(d)  2