A particle of mass $$m$$, charge $$Q$$and kinetic energy$$T$$enters a transverse uniform magnetic field of induction $$\overrightarrow{B}$$. After $3$ seconds the kinetic energy of the particle will be:

The correct option is C

$$T$$

Step 1. Given Data,

Kinetic Energy$=T$

Charge $=Q$

mass of the particle $=m$

Step 2. Formula used,
The work done $$W$$ is
$$W=Fd\cos \theta$$
$$F$$ is the force on the partcle, $$d$$ is the displacement of the particle and $$\theta$$ is the angle between the force and displacement of the particle.
The expression for the work-energy theorem is
$$W=\mathrm{\Delta }K$$
Here,$$W$$ is the work done and $$\mathrm{\Delta }K$$is the change in kinetic energy.

Step 3. Calculating the kinetic energy of the particle,
We know that when a charged particle enters a uniform magnetic field, a magnetic force acts on this charged particle in a direction perpendicular to the direction of motion of the particle.
This force acting in the direction perpendicular to the displacement of the particle causes the charged particle to perform uniform circular motion in the magnetic field. The angle between the magnetic force and the displacement of the particle is $${90}^{\circ }$$.
$$\theta ={90}^{\circ }$$
Calculating the work done on this charged particle in the uniform magnetic field.
$$W=Fd\cos {90}^{\circ }$$
$$\Rightarrow W=Fd\left(0\right)$$
$$\Rightarrow W=0\text{J}$$
Hence, the work done on this charged particle as $$0\text{J}$$ as it is in a uniform circular motion at any interval of time.
Let us determine the kinetic energy of the charged particle 3 seconds after entering the magnetic field.

Substitute $$0\text{J}$$ for $$W$$
$$\left(0\text{J}\right)=\mathrm{\Delta }K$$
$$\therefore \mathrm{\Delta }K=0\text{J}$$
The change in kinetic energy of the charged particle after entering the magnetic field is zero. Therefore, the kinetic energy of the charged particle $3$ seconds after is $$T$$.
Hence, the correct option is C