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Question: A particle of charge \[q\] and mass \[m\] enters normally (at point P) in a region of the magnetic f...

A particle of charge qq and mass mm enters normally (at point P) in a region of the magnetic field with speed vv. It comes out normally from Q after time T as shown in fig. The magnetic field BB is present only in the region of radius RR and is uniform. Initial and final velocities are along the radial direction and they are perpendicular to each other. For this to happen, which of the following expression(s) is/are correct?

& \text{A}\text{. }B=\dfrac{mv}{qR} \\\ & \text{B}\text{. }T=\dfrac{\pi R}{2v} \\\ & \text{C}\text{. }T=\dfrac{\pi m}{2qB} \\\ & \text{D}\text{. none} \\\ \end{aligned}$$
Explanation

Solution

When a charged particle enters a uniform magnetic field, the magnetic force provides centripetal force and the particle executes circular motion inside the magnetic field. For the initial and final velocities to be perpendicular to each other, we will apply the equation for the magnetic force on the particle and determine the expressions for the magnetic field and time period of the motion of the particle.

Formula used:
Magnetic force, F=q(v×B)\overrightarrow{F}=q\left( v\times B \right)
Centrifugal force, Fc=mv2r{{F}_{c}}=\dfrac{m{{v}^{2}}}{r}
The radius of motion of a particle in the magnetic field, r=mvqBr=\dfrac{mv}{qB}

Complete step by step solution:
We know that when a charged particle in motion is subjected to a magnetic field that is perpendicular to its direction of motion, it results in the circular motion of the particle.
Magnetic force on a particle moving of charge qq and mass mm moving with velocity vv when subjected to a perpendicular magnetic field BB the force experienced by the particle is:
F=q(v×B)\overrightarrow{F}=q\left( v\times B \right)
As the cross product of two vectors is always perpendicular to both the vectors, so the resulting force will always be perpendicular to the velocity of the particle.
Thus, it will lead the particle into a circular motion.
F=q(v×B) F=qvBsinθ \begin{aligned} & \overrightarrow{F}=q\left( v\times B \right) \\\ & F=qvB\sin \theta \\\ \end{aligned}
Where θ\theta is the angle between vv and BB
Now, for velocity of particle and magnetic field to be perpendicular,
sinθ=1\sin \theta =1
Or, F=qvBF=qvB
Now, the magnetic force will provide the necessary centripetal force for the circulation motion.
Centripetal force acting on a particle of mass mm moving with velocity vv in a circular path of radius rr is:
Fc=mv2r{{F}_{c}}=\dfrac{m{{v}^{2}}}{r}
Equating the two forces,
qvB=mv2r r=mvqB \begin{aligned} & qvB=\dfrac{m{{v}^{2}}}{r} \\\ & r=\dfrac{mv}{qB} \\\ \end{aligned}
Now, for the given particle, its velocity vector is always perpendicular to the magnetic field, so it will undergo circular motion.
Also, it enters from P and comes out at Q with velocity normal to initial, so making its path a quarter circle.
Radius of this quarter circle will be equal to RR
So, using the above result, r=mvqBr=\dfrac{mv}{qB} for this particle, we have:
R=mvqB B=mvqR \begin{aligned} & R=\dfrac{mv}{qB} \\\ & B=\dfrac{mv}{qR} \\\ \end{aligned}
So, option A is correct.
Now, distance travelled by the particle is one-fourth of the circumference of a complete circle of radius RR
So, distance travelled: 14(2πR)=πR2\dfrac{1}{4}\left( 2\pi R \right)=\dfrac{\pi R}{2}
Speed of particle remains the same as vv
So time taken is distance divided by speed.
T=Ds T=πR2v T=πR2v \begin{aligned} & T=\dfrac{D}{s} \\\ & T=\dfrac{\dfrac{\pi R}{2}}{v} \\\ & T=\dfrac{\pi R}{2v} \\\ \end{aligned}
So option B is correct.
We have: B=mvqRB=\dfrac{mv}{qR}
So, v=BqRmv=\dfrac{BqR}{m}
Substituting v=BqRmv=\dfrac{BqR}{m}in T=πR2vT=\dfrac{\pi R}{2v}
We get,
T=πR2×BqRm T=πRm2BqR T=πm2Bq \begin{aligned} & T=\dfrac{\pi R}{2\times \dfrac{BqR}{m}} \\\ & T=\dfrac{\pi Rm}{2BqR} \\\ & T=\dfrac{\pi m}{2Bq} \\\ \end{aligned}
So, option C is also correct.
Hence, correct options are A, B, and C.

Note: In a circular motion, the velocity of the particle is always perpendicular to the force acting on it. When a charged particle enters the magnetic field, we know that the magnetic force is given as the charge times the cross product of velocity and magnetic field. As a cross-product given the vector which is perpendicular to both the vectors in the cross product, therefore, the magnetic force is always perpendicular to the velocity of the particle and the particle starts executing circular motion.