Question

An electron that has a velocity with x component 2.6 × 106 m/s and y component 4.0 × 106 m/s moves through a uniform magnetic field with x component 0.037 T and y component -0.17 T. (a) Find the magnitude of the magnetic force on the electron. (b) Repeat your calculation for a proton having the same velocity.

Answer 1

Step-by-step explanation:

It is given that,

The horizontal and vertical component of velocity of an electron is:

`v=(2.6i+4j)* 10^6\ m/s`

The magnetic field acting there is given by :

`B=(0.037i-0.17j)\ T`

(a) The magnitude of the magnetic force on the electron is given by :

`F=q(v* B)`

q = e

`F=e(v* B)`

`F=1.6* 10^(-19)* ((2.6i+4j)* 10^6* (0.037i-0.17j))`

`F=-1.6* 10^(-19)* (-0.442\cdot10^(6)-0.1702\cdot10^(6))\ kN`

`F=9.79* 10^(-14)\ kN`

(b) We know that the charge on proton is :

`q=+1.6* 10^(-19)\ C`

The magnetic force as same as for electron but the direction is opposite i.e.

`F=-9.79* 10^(-14)\ kN`

Hence, this is the required solution.