# In electromagnetic induction electric line is always perpendicular to magnetic line?explain briefly?

Arun
25750 Points
5 years ago
The perpendicularity of the electric and magnetic fields follows directly from Maxwell's equations.
I will assume you know basic operations of vector calculus, gradient, divergence and curl.
Assume no charges to be present.

∇×E(r,t) + 1/c ∂_B(r,t) = 0
∇·B(r,t) = 0
∇×B(r,t) − 1/c ∂_E(r,t) = 0
∇·E(r, t= 0

E and B are the electric field strength and the magnetic induction, respectively and c is the velocity of light in free space.

The spatial and time periodicity of the radiation be utilized to write Maxwell’s equations in Fourier transformed form:

q×E(q,ω)− ω/cB(q,ω) = 0
q · B(q, ω) = 0
q×B(q,ω)+ω/cE(q,ω) = 0
q · E(q, ω) = 0
q is a wave vector.

From the third equation we get
B(q, ω)=q×E(q, ω) /(w/c)

Now we take the scalar product with E
E(q, ω)·B(q, ω) = E(q, ω).q×E(q, ω) / (w/c)
but from the first equation we know that  iq · E(q, ω) = 0 ; therefore
E(q, ω)· B(q, ω) = 0

For the scalar product between two vectors to be zero either one of them is the zero vector or they are perpendicular to each other.

Therefore, the electric and magnetic fields are perpendicular.
For a propagating EM wave, the E and B fields are always perpendicular in a homogenous, linear, anisotropic medium. This type of media includes many things like air, water, glass (without stress or tempering).
However, in inhomogenous, non-linear, or isotropic media, the E and B fields may not be perpendicular, e.g. in a crystal (which is isotropic).