Question

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?

Answer 1

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 )

= 42.33^o

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

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 )

= 24.88^o

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

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 )

= 0.803^o