Question

In an electromagnetic wave,
$E=1.2\sin (2\times {10}^{-6}t-kx)N/C$.
Find value of intensity of magnetic field.
$4\times {10}^{-8}field.$

• A. $4\times {10}^{-8}A/m$
• B. $0.1\times {10}^{-2}A/m$
• C. $0.4\times {10}^{-8}A/m$
• D. $\frac{{10}^{-2}}{\pi }A/m$

Answer 1

Answer: (D)

$\frac{{10}^{-2}}{\pi }A/m$

Given: In an electromagnetic wave, $E=1.2\sin (2\times {10}^{6}t-kx)N/C$

To find the value of intensity of magnetic field

Solution:

we know the equation of the electromagnetic wave is

$E=E_0\sin (kx-\omega t)$

Comparing this with the given equation,

$E_0=1.2$

The value of intensity of magnetic field

$B_0=\frac{E_0}{c}⟹B_0=\frac{1.2}{3\times {10}^8}⟹B_0=0.4\times {10}^{-8}A/m$