Question

A 0.40 *μ*C 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 *μ*C

= 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 *μ*C

= 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 *μ*C

= 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}**

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