A particle of mass 4.5 × 10-8 kg and charge +5.4 μC is traveling due east. It enters perpendicularly a magnetic field whose magnitude is 2.7 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?

0.00970 s

Explanation:

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

Force due to magnetic field = qvB sin θ

q = charge on the particle = 5.4 μC

v = velocity of the charge

B = magnetic field strength = 2.7 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.

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

mvw = qvB

mw = qB

w = (qB/m) = (5.4 × 10⁻⁶ × 2.7)/(4.5 × 10⁻⁸)

w = 3.24 × 10² rad/s

w = 324 rad/s

w = (angular displacement)/time

Time = (angular displacement)/w

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

Time = (π/324) = 0.00970 s

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electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j