A particle with a charge of +8.4 µC and a speed of 35 m/s enters a uniform magne

Question

A particle with a charge of +8.4 µC and a speed of 35 m/s enters a uniform magnetic field whose magnitude is 0.29 T. For each of the cases in the drawing, find the magnitude and direction of the magnetic force on the particle.

A particle with a charge of +8.4 MuC and a speed of 35 m/s enters a uniform magnetic field whose magnitude is 0.29 T. For each of the cases in the drawing, find the magnitude and direction of the magnetic force on the particle. (a) magnitude 1 N direction 2 ---Select--- out of the screen downwards upwards into the screen (b) magnitude 3 N direction 4 ---Select--- out of the screen into the screen upwards downwards (c) magnitude 5 N direction 6 ---Select--- into the screen upwards downwards out of the screen

Explanation / Answer

The velocity vector, Magnetic field vector be V and B respectively. then the force in both magnitude and direction is given by
F = q (v X B) = qvB sin ( is angle between v and B) [Vector cross product]

Here q= +8.4 µC = 8.4 x 10-6 C

|v|=magnitude of v = 35 m/s

B = 0.29T

a) F= qvB sin

=   8.4 x 10-6 C x 35 m/s x 0.29 x sin30

= 4.263 x 10-5 N ,into the screen downwards (into the screen)

b) F= qvB sin

=   8.4 x 10-6 C x 35 m/s x 0.29 x sin90

= 8.526 x 10-5 N ,into the screen downwards (into the screen)

a) F= qvB sin

=   8.4 x 10-6 C x 35 m/s x 0.29 x sin150

= 4.263 x 10-5 N ,into the screen downwards (into the screen)

Howw to find direction :

Place your right hand fingers alond V and curl them into the direction of B, the direction of your thum gives the direction of the vector product v x B.

Hope you got it :)

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