Question

A light beam travelling in the x-direction is described by the electric field $E_y=300v/m\sin \left(t-\frac{x}{c}\right)$. An electron is constrained to move along the y-direction with a speed of $2.0\times {10}^{7}m/s.$ Find the maximum magnetic force on the electron.

• A. $2\times {10}^{-18}$
• B. $2.5\times {10}^{-18}$
• C. $3\times {10}^{-18}$
• D. $3.2\times {10}^{-18}$

Answer 1

The correct option is D $3.2\times {10}^{-18}$

As we know that,

$B_{\circ }=\frac{E_{\circ }}{C}$

Where,

$B_{\circ }$ = Max. Magnetic Field

$E_{\circ }$ = Max. Electric Field

C= speed of light

It is given that,

$E_{\circ }$= 300 V/m

So, $B_{\circ }$=$\frac{300}{3\times {10}^8}$ = ${10}^{-6}$ T

Now, maximum magnetic force= qv$B_{\circ }$

= $1.6\times {10}^{-19}\times 2\times {10}^7\times {10}^{-6}$

=$3.2\times {10}^{-18}$ N