Question

An electron that has a velocity with x component 1.6 × 106 m/s and y component 2.4 × 106 m/s moves through a uniform magnetic field with x component 0.025 T and y component -0.16 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

Answer:

(a) 5.056 x 10^-14 N

(b) 5.056 x 10^-14 N

Explanation:

X component of velocity of electron is 1.6 × 10^6 m/s

Y component of velocity of electron is 2.4 × 10^6 m/s

X component of magnetic field is 0.025 T

Y component of magnetic field is  -0.16 T

charge on electron, q = - 1.6 x 10^-19 C

Write the velocity and magnetic field in the vector forms.

$\overrightarrow{v}=1.6\times 10^{6}\widehat{i}+2.4\times 10^{6}\widehat{j}$

$\overrightarrow{B}=0.025\widehat{i}-0.16\widehat{j}$

The force on the charge particle when it is moving in the magnetic field is given by

$\overrightarrow{F}=q\left ( \overrightarrow{v}\times \overrightarrow{B} \right )$

(a) Force on electron is given by

$\overrightarrow{F}=-1.6\times 10^{-19}\left ( 1.6\times 10^{6}\widehat{i}+2.4\times 10^{6}\widehat{j} \right )\times \left ( 0.025\widehat{i}-0.16\widehat{j} \right )$

$\overrightarrow{F}=5.056\times 10^{-14}\widehat{k}$

Magnitude of force is 5.056 x 10^-14 N.

(b) Force on a proton is given by

$\overrightarrow{F}=1.6\times 10^{-19}\left ( 1.6\times 10^{6}\widehat{i}+2.4\times 10^{6}\widehat{j} \right )\times \left ( 0.025\widehat{i}-0.16\widehat{j} \right )$

$\overrightarrow{F}=-5.056\times 10^{-14}\widehat{k}$

Magnitude of force is 5.056 x 10^-14 N.

Thus, the magnitude of force remains same but the direction of force is opposite to each other.

Explanation: