Question

How fast must proton be moving if it is to follow a circular path with radius 2cm in magnetic field of 0.7T ?

Answer 1

The proton will be moving if it is to follow a circular path with radius 2cm in magnetic field of 0.7T  is 1.34 x 10⁷  m/s.

What is magnetic field?

It is the region of space where a charge experiences the magnetic force when it enters the field.

While moving in the circular path, centripetal force is equal to the magnetic force.

Fm = Fc

q x v x B = m x v² / r

v= qBr /m

where mass of proton m =1.67262 × 10−27 kg, charge of proton q =1.6 x 10⁻¹⁹ C and magnetic field B =0.7T.

Then, velocity is

v = 1.6 x 10⁻¹⁹ x 0.7 x 0.02 / 1.67262 × 10⁻²⁷

v = 1.34 x 10⁷ m/s.

Thus, the proton will be moving if it is to follow a circular path with radius 2cm in magnetic field of 0.7T  is 1.34 x 10⁷  m/s.

Learn more about magnetic field.

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