A charged particle is placed in an external magnetic field and it is moving in a circular path of radius 26m.The magnetic force acting on the particle is 16N.What is the kinetic energy of the particle
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Answer:
208 Joules
Explanation:
The radius of the circular path the charge moves, r = 26 m
The magnetic force acting on the charge particle, F = 16 N
Centripetal force,
= m·v²/r
Kinetic energy, K.E. = (1/2)·m·v²
Where;
m = The mass of the charged particle
v = The velocity of the charged particle
r = The radius of the path of the charged particle
Whereby the magnetic force acting on the charge particle = The centripetal force, we have;
F =
= m·v²/r = 16 N
(1/2) × r ×
= (1/2) × r × m·v²/r = (1/2)·m·v² = K.E.
∴ (1/2) × r ×
= (1/2) × 26 m × 16 N = = (1/2)·m·v² = K.E.
∴ 208 Joules = K.E.
The kinetic energy of an particle moving in the circular path, K.E. = 208 Joules.