A uniform 1.5 T magnetic field points north. In an electron moves vertically downward (toward the ground) with a speed of 2.5 x 10^7 m/s through this field, what force (magnitude and direction) will act on it?

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We can solve this problem using the expression:

`F = q*v*B`

Where:

F = force acting on the particle

q = charge of the electron= 1.6022 × 10^-19 coulombs

v = velocity of the moving particle = 2.5 x 10^7 m/s

B = magnetic field = 1.5 T

By substituting the values we can no have:

`F = q*v*B`

`F = (1.6022x10^-19)*(2.5 x 10^7)*(1.5)`

`F = 6.0x10^-12 N`

The direction of the force can be determined using the right hand rule. So for this problem, the direction is west.

**Answer: 6.0x10^-12 N, West**

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