Answer :
Answer:
v = 5.75 x 10⁶ m/s
Explanation:
The radius (r) of the circular orbit taken by a charged particle is related to its speed perpendicular to a magnetic field of strength B, and is given by
r = [tex]\frac{mv}{qB}[/tex] --------------(i)
Where,
q = charge of the particle
m = mass of the particle
Making v subject of the formula in equation (i) above gives
v = [tex]\frac{qBr}{m}[/tex] -------------------(ii)
Given;
r = 20cm = 0.2m
B = 0.3T
v = unknown
q = charge of proton = 1.6 x 10⁻¹⁹ C
m = mass of the proton = 1.67 x 10⁻²⁷kg
Substitute the values of m, q, B and r into equation (ii) above to get;
v = [tex]\frac{1.6 * 10^{-19} * 0.3 * 0.2} {1.67*10^{-27} }[/tex]
Solving for v gives:
v = 5.75 x 10⁶ m/s
Therefore, the velocity of the proton is 5.75 x 10⁶ m/s