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  • The electric field in a plane electromagnetic wave is given by

Q.2) The electric field in a plane electromagnetic wave is given by

$$
E_z=60 \cos \left(5 x+1.5 \times 10^9 t\right) \ V / \mathrm{m}
$$


Then expression for the corresponding magnetic field is (here, subscripts denote the direction of the field) :

A) $B_y=60 \sin \left(5 x+1.5 \times 10^9 t\right) T$

B) $B_y=2 \times 10^{-7} \cos \left(5 x+1.5 \times 10^9 t\right) T$

C) $B_x=2 \times 10^{-7} \cos \left(5 x+1.5 \times 10^9 t\right) T$

D) $B_z=60 \cos \left(5 x+1.5 \times 10^9 t\right) T$

 

Answers (1)

best_answer

$$
B y=2 \times 10^{-7} \cos \left(5 x+1.5 \times 10^9 t\right) T
$$


Solution:
In an electromagnetic wave, the electric and magnetic fields are perpendicular and the magnetic field amplitude is $\mathrm{B}=\mathrm{E} / \mathrm{c}$. Given $\mathrm{E}_0=60 \mathrm{~V} / \mathrm{m}$ and $\mathrm{c}=3 \times 10^8 \mathrm{~m} / \mathrm{s}$,

$$
\mathrm{B}_0=60 /\left(3 \times 10^8\right)=2 \times 10^{-7} \mathrm{~T} .
$$


So the magnetic field has the same cosine form and is in the $y$-direction.

 

Posted by

avinash.dongre

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