A hypothetical material has a relative permeability and relative permittivity of 3.0 and 4.0, respectively.

Question

A hypothetical material has a relative permeability and relative permittivity of 3.0 and 4.0, respectively.

a. What is the speed of light in this material?

b. What is the value of the magnetic field if the electric field is 3.5 N/C?

Answer 1

To find the speed of light in the material, we can use a formula

v = c / √KK_m

After substitution of the given values we get

v = 3×10⁸ m/s / √3·4 = 8,7×10⁷ m/s.

Assuming that the maximum value of electric field is given, we can find magnetic field using the formula

B_max = E_m / c

Or

B_max = 3.5 N/C / 3×10⁸ m/s = 1.2×10 ^-8 T