Question

An electron that has an instantaneous velocity of ???? = 2.0 × 106 m ???? ???? + 3.0 × 106 m ???? ???? is moving through the uniform magnetic field ???? = 0.030T ???? − 0.15T ???? . (a) Find the force on the electron due to the magnetic field (b) Repeat your calculation for a proton having the same velocity.

Answer 1

Explanation:

It is given that,

Velocity of the electron, $v=(2\times 10^6i+3\times 10^6j)\ m/s$

Magnetic field, $B=(0.030i-0.15j)\ T$

Charge of electron, $q_e=-1.6\times 10^{-19}\ C$

(a) Let $F_e$ is the force on the electron due to the magnetic field. The magnetic force acting on it is given by :

$F_e=q_e(v\times B)$

$F_e=1.6\times 10^{-19}\times [(2\times 10^6i+3\times 10^6j)\times (0.030i-0.15j)]$

$F_e=-1.6\times 10^{-19}\times (-390000)(k)$

$F_e=6.24\times 10^{-14}k\ N$

(b) The charge of electron, $q_p=1.6\times 10^{-19}\ C$

The force acting on the proton is same as force on electron but in opposite direction i.e (-k). Hence, this is the required solution.