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

Step-by-step explanation:

It is given that,

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

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

Charge of electron,
`q_e=-1.6* 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* B)`

`F_e=1.6* 10^(-19)* [(2* 10^6i+3* 10^6j)* (0.030i-0.15j)]`

`F_e=-1.6* 10^(-19)* (-390000)(k)`

`F_e=6.24* 10^(-14)k\ N`

(b) The charge of electron,
`q_p=1.6* 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.