Question

proton enters a perpendicular magnetic field of intensity ‘B’ with speed ‘v’. The

proton moves along a circular path of radius ‘R’. If a deuteron enter in to the same magnetic

field with same speed, the radius of its circular trajectory will be?

(a) R

(b) Less than R

(c) More than R

(d) cannot be said.​

Answer 1

(b) Less than R.

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Answer 2

Given: A proton enters a magnetic field on intensity B

            Speed of proton = v m/s

            Radius of circle traced by proton = R m

To Find: Radius of a deuteron with same speed if it enters the same magnetic field.

Solution:

• Force experienced by a particle in a magnetic field is given by:

                                F = qvB sinФ

where, q is charge of the particle

            v is velocity of the particle

            B is intensity of magnetic field

      and Ф is angle made by particle while entering field

• Since proton moves in a circular path, we can say that magnetic force provides centripetal acceleration to the proton.

                                    ∴ qvB = $\frac{mv^{2} }{r}$        (1)         (here Ф= 90° and sin90=1)

           Substituting given values in (1)

                                 qvB = $\frac{mv^{2} }{R}$                    (centripetal force = $\frac{mv^{2} }{r}$)

                         ⇒      qB = $\frac{mv}{R}$

                        ⇒       R = $\frac{mv}{qB}$                        (2)

• Now for a deuteron, charge is of + 1 q( q is charge on proton) and mass is twice that of a proton due to presence of a neutron.

Substituting these values in (2) we get:

                                   ⇒  R' = $\frac{2mv}{qB}$ = 2R

Hence radius of circular path transcribed by deuteron in the same magnetic field with same intensity will be 2R.

So, option (c) more than R is correct.