Question

What is the magnetic force acting on an electron if its speed is 3.0 × 106 meters/second and the direction is perpendicular to a magnetic field of 0.020 teslas? The value of q = -1.6 × 10-19 coulombs.

Answer 1

Answer : The magnetic force acting on an electron is, $F=9.6\times 10^{-15}\ N$.

Explanation :

It is given that,

Speed of electron, $v = 3\times 10^6\ m/s$

Magnetic field, B = 0.020 T

Charge on electron, $q=-1.6\times 10^{-19}\ C$

The formula of magnetic force,

$F =q(v\times B)$

$F=qvBsin\ 90$

$F=1.6\times 10^{-19}\ C \times 3\times 10^6\ m/s \times 0.020 T$

$F=9.6\times 10^{-15}\ N$

Hence, this is the required solution.

Answer 2

Magnetic force = charge × speed × magnetic field

Fm = (1.6x10^-19)(3x10^6)(0.02)

Fm = 9.60x10^-15 N

^That is the magnitude of the magnetic force. If you want the direction, you will need to know what direction the speed and magnetic field are. Then you can use the left hand straight fingered rule to determine the direction of the force.