A particle of mass 7.4 × 10-8 kg and charge +8.6 μC is traveling due east. It enters perpendicularly a magnetic field whose magnitude is 3.2 T. After entering the field, the particle completes one-half of a circle and exits the field traveling due west. How much time does the particle spend in the magnetic field?

Time spent by the charge in the magnetic field = 0.008445 s = (8.445 × 10⁻³) s

Explanation:

The force exerted on the charge due to magnetic field = the centripetal force that causes the charge to move in a circular motion.

Force due to magnetic field = qvB sin θ

q = charge on the particle = 8.6 μC

v = velocity of the charge

B = magnetic field strength = 3.2 T

θ = angle between the velocity of the charge and the magnetic field = 90°, sin 90° = 1

F = qvB

Centripetal force responsible for circular motion = mv²/r = mvw

where w = angular velocity.

mvw = qvB

mw = qB

w = (qB/m) = (8.6 × 10⁻⁶ × 3.2)/(7.4 × 10⁻⁸)

w = 3.72 × 10² rad/s

w = 372 rad/s

w = (angular displacement)/time

Time = (angular displacement)/w

Angular displacement = π rads (half of a circle; 2π/2)

Time = (π/372) = 0.008445 s

-4 on

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