A current-carrying closed Loop in the form of a right isosceles triangle ABC is placed in xy plane in a uniform magnetic field acting along the y-axis as shown in the figure. If the magnetic force on the arm AC is $\overrightarrow{F}$, then force on the arm BC and AB are respectively

1. zero, zero

2. $-\overrightarrow{F}<mspace\; width="1em"></mspace>and<mspace\; width="1em"></mspace>zero$

3. $\overrightarrow{F}<mspace\; width="1em"></mspace>and<mspace\; width="1em"></mspace>-\overrightarrow{F}$

4. $\overrightarrow{F}<mspace\; width="1em"></mspace>and<mspace\; width="1em"></mspace>zero$

A uniform electric field and a uniform magnetic field are acting in the same direction in a certain region. If a proton is projected in this region such that its velocity is pointed along the direction of fields, then the proton :

1.  will turn towards the left of the direction of motion

2.  will turn towards the right of the direction of motion

3.  will speed up

4.  will slow down

When a proton is released from rest in a room, it starts with an initial acceleration ${\mathrm{a}}_0$ towards the east. When it is projected towards the north with a speed ${\mathrm{v}}_0$, it moves with initial acceleration $3{\mathrm{a}}_0$ towards east. The electric and magnetic fields in the room are -

1.  $\frac{{\mathrm{Ma}}_0}{\mathrm{e}}\mathrm{west},<mspace\; width="1em"></mspace>\frac{{\mathrm{Ma}}_0}{{\mathrm{ev}}_0}\mathrm{up}$

2.  $\frac{{\mathrm{Ma}}_0}{\mathrm{e}}\mathrm{west},<mspace\; width="1em"></mspace>\frac{2{\mathrm{Ma}}_0}{{\mathrm{ev}}_0}\mathrm{down}$

3.  $\frac{{\mathrm{Ma}}_0}{\mathrm{e}}\mathrm{east},<mspace\; width="1em"></mspace>\frac{2{\mathrm{Ma}}_0}{{\mathrm{ev}}_0}down$

4.  $\frac{{\mathrm{Ma}}_0}{\mathrm{e}}\mathrm{east},<mspace\; width="1em"></mspace>\frac{2{\mathrm{Ma}}_0}{{\mathrm{ev}}_0}up$

When an electron enters perpendicularly in a magnetic field with velocity v, time period of its revolution is T. If it enters in the same magnetic field with a velocity 2v, then its time period will be:

1.  2T

2.  4T

3.  $\frac{\mathrm{T}}{2}$

4.   T

A positively charged particle descends vertically. In which direction will the horizontal component of the earth's magnetic field deflect?

1. east

2. west

3. north

4. south

The figure shows a particle of charge q and mass m moving with velocity v along the x-axis enters a region of the uniform magnetic field. The minimum value of v so that the charge q is deflected by an angle $30°$ is

1.  $\frac{2\mathrm{qB}\left(\mathrm{b}<mspace\; width="1em"></mspace>-<mspace\; width="1em"></mspace>\mathrm{a}\right)}{\mathrm{m}}<mspace\; width="1em"></mspace>$

2.  $\frac{\mathrm{qB}\left(\mathrm{b}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>\mathrm{a}\right)}{2\mathrm{m}}<mspace\; width="1em"></mspace>$

3.  $\frac{\mathrm{qB}\left(\mathrm{b}<mspace\; width="1em"></mspace>-<mspace\; width="1em"></mspace>\mathrm{a}\right)}{\mathrm{m}}<mspace\; width="1em"></mspace>$

4.  $\frac{\mathrm{qBb}}{2\mathrm{m}}$

An electron is traveling along the x-direction. It encounters a magnetic field in the y-direction. Its subsequent motion will be:

1.  Straight-line along the x-direction

2.  A circle in the xz-plane

3.  A circle in the yz-plane

4.  A circle in the xy-plane

The dots in the figure depict a magnetic field that is perpendicular to the plane of the paper and emanates from it. The trajectory of a particle in the plane of the paper is depicted by the curve \(ABC\). What exactly is the particle?

The electric charge in uniform motion produces :

An electron and a proton enter a magnetic field perpendicularly. Both have the same kinetic energy. Which of the following is true:

1. Trajectory of electron is less curved

2. Trajectory of proton is less curved

3. Both trajectories are equally curved

4. Both move on a straight-line path