Question

# A 0.40 μC particle moves with a speed of 18 m/s through a region where the...

A 0.40 μC particle moves with a speed of 18 m/s through a region where the magnetic field has a strength of 0.99 T

At what angle to the field is the particle moving if the force exerted on it is 4.8×10−6N?

At what angle to the field is the particle moving if the force exerted on it is 3.0×10−6N?

At what angle to the field is the particle moving if the force exerted on it is 1.0×10−7N?

q = 0.40 μC

= 0.4 X 10-6 C

v = 18 m/s

B = 0.99 T

F = 4.8 X 10−6 N

using

F = q v B sin

= sin-1 ( F/qvB )

= sin-1 ( 4.8 X 10−6 / 0.4 X 10-6 X 18 X 0.99 )

= 42.33o

--------------------------------------------------------------------------------

q = 0.40 μC

= 0.4 X 10-6 C

v = 18 m/s

B = 0.99 T

F = 3 X 10−6 N

using

F = q v B sin

= sin-1 ( F/qvB )

= sin-1 ( 3 X 10−6 / 0.4 X 10-6 X 18 X 0.99 )

= 24.88o

--------------------------------------------------------------

q = 0.40 μC

= 0.4 X 10-6 C

v = 18 m/s

B = 0.99 T

F = 1 X 10−7 N

using

F = q v B sin

= sin-1 ( F/qvB )

= sin-1 ( 1 X 10−7 / 0.4 X 10-6 X 18 X 0.99 )

= 0.803o

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