Q.220. Motion of proton in a magnetic field.

Question 220.
A proton is projected with a speed of 500 m/s in a uniform magnetic field of 5T. If the angle between velocity & force vector is 45°.  Find the pitch of its helical path.
 Answer 220.
Speed parallel and perpendicular to the magnetic field
= 500cos(45°)
= 250sqrt(2) m/s.
Radius of the helical curve is given by
qvB = mv^2/r
=> r = mv/qB
=> r
= (1.67262158 × 10-27)*(250sqrt(2)) / (1.6x10^-19 * 5) m
= 7.392 x 10^-8 m

 Pitch of the helical path of the proton,
= 2πr/(vcos45°) * vsin45
= 2πr
= 2 x 3.14159 x 7.392 x 10^-8 m
= 4.64 x 10^-7 m.

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