Question

A proton is moving at a speed of 3.2 × 106 m/s at an angle of 80° to a uniform magnetic field of strength 1.5 × 10-4 T. What is the distance moved by the proton along the spiral path as it completes one revolution? (This distance is called the "pitch" of the helix.)

Answer 1

For a positive charge entering a magnetic field with an angle the path followed is following:

The radius of the circular motion is given by the equation

r=mvsinθ/qB

and the pitch of the helix is

p=2πmvcosθ/qB

Putting values from question

Angle = 80°

v = 3.2×10⁶ m/s

B = 1.5×10^-4 T

m = 1.67 × 10^-27 Kg

q = 1.60 × 10^-19 C

Assigning above values in equation we get,

p = (2×3.14×1.67×10^-27×3.2×10⁶×cos80)/1.6×10^-19×1.5×10^-4

p = 2.427 × 10² m

Hence the pitch of helix of proton will be equal to 242.79 m.

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