Answer to Question #185055 in Mechanics | Relativity for jay

Question

A uniform magnetic field, B= 3.0 x 10^-4T in the +x direction. A proton (q=-+e) shoots through the field in the +y direction with a speed of 5.0 x 10⁶m/s.

(a) Find the magnitude and direction of the force on the proton.

(b) Repeat with the proton replaced by an electron.

Answer 1

To be given in question

Magnetic force (B)=$3\times10^{-4}T$

Charge of proton (q)=$+1.6\times10^{-19}C$

Velocity (v)=$5\times10^6m/sec$

$\theta=90°$

To be asked in question

(a) magnetic force (for proton)=?

(b) magnetic force (for electron)=?

We know that

(a) magnetic force $( \overrightarrow F)=q(\overrightarrow{v}\times \overrightarrow{B})$

$F=qvBsin\theta$$\hat{n}$

sin90°=1

Put values

$F =qvB\hat{n}$

$\hat{n}$ direction is $( \overrightarrow v\times \overrightarrow B)$ direction

$F={1.6\times10^{-19}\times5\times10^6\times3\times10^4}\hat{ n}$ $F=2.4\times10^{-8} (\hat{n})N$

$\hat{n}$ direction is $-\hat{k}$ direction

Magnitude of magnetic force

$F=2.4\times 10^{-8}$$-(\hat{k})$

Magnetic force direction $-\hat{k}$

Negative (-) z- axis

(b) for electron magnetic force

$\overrightarrow F=q( \overrightarrow v\times \overrightarrow B)$

$F=qvBsin\theta(\hat{n})$

$\hat{n}$ direction is $( \overrightarrow v\times \overrightarrow B)$ direction

Put value

$F={-1.6\times10^{-19}\times5\times10^6\times3\times10^4}(\hat{n})$ $F=-2.4\times10^{-8}(\hat n)N$

$\hat{n}$ direction is $-\hat{k}$ But charge is$q=-1.6\times10^{-19}C$ then direction is$-(-\hat n)=\hat{n}$

Resultant $\hat{n}$ direction is$+\hat{k}$ direction

$F=2.4\times 10^{-8}(\hat n )N$

Magnitude of magnetic force

$F=2.4\times 10^{-8}N$ $(+\hat{k})$

magnetic force direction

$(+\hat{k})$

Positive (+ )z-axis