Question

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:
A. $$T$$
B. $$4T$$
C. $$3T$$
D. $$2T$$

Answer 1

Hint:Use the formula for the work done on an object by the force in order to displace it. Also use the expression for work energy theorem which gives the relation between the work and change in kinetic energy of an object. First determine the work done by the magnetic force on the charged particle when it enters the magnetic field and then determine the change in kinetic energy of the particle to calculate kinetic energy after 3 seconds.

Formula used:
The work done $$W$$ is given by
$$W=Fd\cos \theta$$ …… (1)
Here, $$F$$ is the force on the object, $$d$$ is the displacement of the object and $$\theta$$ is the angle between the force and displacement of the object.
The expression for work energy theorem is
$$W=\mathrm{\Delta }K$$ …… (2)
Here, $$W$$ is the work done and $$\mathrm{\Delta }K$$ is the change in kinetic energy.

Complete step by step answer:
We have given that the mass of the particle is $$m$$, charge of the particle is $$Q$$ and kinetic energy of the particle is $$T$$ and this particle enters a transverse uniform magnetic field. We have asked to calculate the kinetic energy of the particle 3 seconds after it enters the magnetic field. We know that when a charged particle enters in an 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 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 $${90}^{\circ }$$.
$$\theta ={90}^{\circ }$$

Let us calculate the work done on this charged particle in the uniform magnetic field.
Substitute $${90}^{\circ }$$ for $$\theta$$ in equation (1).
$$W=Fd\cos {90}^{\circ }$$
$$\Rightarrow W=Fd\left(0\right)$$
$$\Rightarrow W=0\text{J}$$
Hence, the work done on this charged particle is $$0\text{J}$$ as it is in the uniform circular motion.

Let us determine the kinetic energy of the charged particle 3 seconds after entering the magnetic field. Substitute $$0\text{J}$$ for $$W$$ in equation (2).
$$\left(0\text{J}\right)=\mathrm{\Delta }K$$
$$\therefore \mathrm{\Delta }K=0\text{J}$$
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 A.

Note:One can also solve the same question by another method. One can show that the kinetic energy of a particle depends only on the magnitude of mass and velocity of the particle and when a particle enters a uniform magnetic field, there is change in only direction of velocity of the particle. The mass and magnitude of velocity remains the same. Hence, the kinetic energy of the particle also remains the same at any time.