Question

What must be the magnitude of a uniform electric field if it is to have the same energy density as that possessed by a 0.50 T magnetic field?

Answer 1

E = 1.50 ×
`10^(8)` V/m

Step-by-step explanation:

given data

B = 0.50 T

solution

we know that energy density by the magnetic field is express as

`\mu _b = (B)/(2\mu _o)` ...............1

and

energy density due to electric filed is

`\mu _e = (\epsilon _o E^2)/(2)` ...............2

and here
`\mu _b = \mu _ e`

so that

E =
`(B)/(√(\mu _o * \epsilon _o))` ...................3

put here value and we get

`E = \frac{0.50}{\sqrt{4\pi * 10^(-7) * 8.852 * 10^(-12)}}`

E = 3 ×
`10^(8)` × 0.50

E = 1.50 ×
`10^(8)` V/m