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A proton enters in a uniform magnetic field of 2.0mT at an angle $60^{\circ}$ with the magnetic field with speed 10m/s. Find the pitch of the path.

Answer
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Hint: When a charged particle goes in a region of magnetic field, two different forces act on it. First one is due to the magnetic field and the other force is centripetal force. The velocity of a particle is constant in the direction parallel to the magnetic field and the motion is accelerating in the direction perpendicular to the magnetic field. We will find the expression for velocity, time period and pitch of the helical path described by the particle.

Formula used:
Magnetic force $P = \dfrac{{2\pi m}}{{qB}}{v_0}$

Complete answer:
Suppose that a charged particle of mass m moves in a circular orbit of radius r, with a constant speed v.
when a charge particle enters in a uniform magnetic field at an angle with some velocity, path of charge must be helix because perpendicular component of velocity creates circular motion whereas parallel component of velocity creates linear motion.

Now the pitch of the helical path is the horizontal distance between two consecutive circles.
i.e., pitch of path = parallel component of velocity × time period
$ = ({v_0}\cos \theta )T$
$ \Rightarrow P = ({v_0}\cos \theta )\dfrac{{2\pi m}}{{qB}}$
$ \Rightarrow P = 10 \times \cos {60^ \circ } \times (\dfrac{{2\pi \times 1.6 \times {{10}^{ - 27}}}}{{1.6 \times {{10}^{ - 19}} \times 2 \times {{10}^{ - 3}}}})$
$ \Rightarrow P = 50\pi \mu m$

Therefore, the pitch is $50~{\pi}{\mu}~m$.

Note: Motion of charged particles is accelerating, in circular motion, in the direction perpendicular to the magnetic field. But, in the direction parallel to the magnetic field, the particle moves with a fixed speed because the magnetic field applies force only in the direction perpendicular to it. Thus, the force can’t do work on the particle, meaning that there is no change in its kinetic energy and therefore, the particle moves with a fixed speed in the direction parallel to the magnetic field.