In a space laboratory, an object of mass m = 10 -3 kg and charge q = 2 x 10 -5 C
ID: 1985805 • Letter: I
Question
In a space laboratory, an object of mass m = 10-3 kg and charge q = 2 x 10-5 C moves in a circular orbit of radius r = 102 m about a fixed charge Q = -10-4 C. There is a uniform magnetic field B = 4 N-s/C-m down into the plane of the orbit.
(INCLUDE DIGRAM and show all work)
Find:
(A) the magnitude and direction of the electric force acting on the object of mass m
(B) themagnetic force on the object in terms of its velocity v
(C) the direction of the magnetic force if
(i) the object moves in a clockwise direction as you look down at the orbit and (ii) if the object moves in the opposite direction.
(D) Use Newton’s second law of motion to determine for what velocity v the object will move in a circle of radius 102 m.
Explanation / Answer
Given
Mass of object m = 10-3 kg
Charge on object q = 2 * 10-5 C
Fixed charge Q = 10-4 C
radius of orbit r = 100 m
Magnitude of uniform magnetic field is B = 4 T
a) Magnitude of electric force
F = k Q q / r2
= ( 9 *109 N m 2 / C 2 ) ( 10-4 C)( 2 * 10-5 C ) / ( 100 m )2
= 0.0018 N
Electric field lines moves from positive charge to negative charge. so the direction of electric force is towards fixed charge.
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b)
Magnetic force on the object is
F B = B q v
= ( 4 T ) ( 2 * 10-5 C ) v
= (8*10-5 ) v
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c)
According to flemings left hand rules the direction of magnetic force is into the page ( down ward direction ) when the object moves in clock wise direction.
The direction of magnetic force is up ward , when the object moves in anti clock wise direction .
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d)
Here the particle in moving in circular motion, force acting on the object is equal to centripetal force.
m v 2 / r = B q v
Velocity of the object is
v = B q r / m
= ( 4 T ) ( 2 * 10-5 C ) ( 100 m ) / ( 10-3 kg )
= 8 m/s