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

Explanation:

It is given that,

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

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

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 :

$F=q(v\times B)$

q = e

$F=e(v\times B)$

$F=1.6\times 10^{-19}\times ((2.6i+4j)\times 10^6\times (0.037i-0.17j))$

$F=-1.6\times 10^{-19}\times (-0.442\cdot10^{6}-0.1702\cdot10^{6})\ kN$

$F=9.79\times 10^{-14}\ kN$

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

$q=+1.6\times 10^{-19}\ C$

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

$F=-9.79\times 10^{-14}\ kN$

Hence, this is the required solution.