Question

An electron that has a velocity with x component 1.6 × 10^6 m/s and y component 2.6 × 10^6 m/s moves through a uniform magnetic field with x component 0.024 T and y component -0.14 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

(a)

`\overrightarrow{F}=4.58*10^(-14)\widehat{K}N`

(b)
`\overrightarrow{F}=- 4.58*10^(-14)\widehat{K}N`

Step-by-step explanation:

Vx = 1.6 x 10^6 m/s

Vy = 2.6 x 10^6 m/s

Bx = 0.024 T

By = - 0.14 T

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

charge of proton, q = 1.6 x 10^-19 C

(a) Force on electron is given by

`\overrightarrow{F}=q(\overrightarrow{V}* \overrightarrow{B})`

Substituting the values

`\overrightarrow{F}=-1.6*10^(-19){\left ( 1.6* 10^(6)\widehat{i}+2.6 * 10^(6)\widehat{j} \right )* \left ( 0.024\widehat{i}-0.14\widehat{j} \right )}`

`\overrightarrow{F}=4.58*10^(-14)\widehat{K}N`

(b) Force on proton is given by

`\overrightarrow{F}=q(\overrightarrow{V}* \overrightarrow{B})`

Substituting the values

`\overrightarrow{F}=1.6*10^(-19){\left ( 1.6* 10^(6)\widehat{i}+2.6 * 10^(6)\widehat{j} \right )* \left ( 0.024\widehat{i}-0.14\widehat{j} \right )}`

`\overrightarrow{F}=- 4.58*10^(-14)\widehat{K}N`