Question

$\textbf{\underline{\underline{Question:-}}}$
The magnitude of electric field E at a point in free space is 300 V/m. Find the magnitude of magnetic field B at this point. Velocity of light is 3 × 10^8 m/s.

______Thank you ☺​

Answer 1

hola user...

here, formula to be applied is

E/ B = C

where, E = electric field

B = magnetic field

C = speed of light

now, put the given values..
uh will get ur answer..

B = 300/3*10^8

B = 10^-6T

Answer 2

Answer:

$\boxed{Your\:Solution}$

Explanation:

$Velocity\:of\:light  =  3  X  10^{8}  m/s$

$C= \frac{E}{B}$

$B=\frac{E}{C}$

$B=\frac{300}{3}X 10^{-8}$

$B=10^{-6}T$

$\boxed{Hope\:it\:Helps}$