JEE MAIN - Physics (2022 - 24th June Morning Shift - No. 13)
A plane electromagnetic wave travels in a medium of relative permeability 1.61 and relative permittivity 6.44. If magnitude of magnetic intensity is 4.5 $$\times$$ 10$$-$$2 Am$$-$$1 at a point, what will be the approximate magnitude of electric field intensity at that point?
(Given : Permeability of free space $$\mu$$0 = 4$$\pi$$ $$\times$$ 10$$-$$7 NA$$-$$2, speed of light in vacuum c = 3 $$\times$$ 108 ms$$-$$1)
16.96 Vm$$-$$1
2.25 $$\times$$ 10$$-$$2 Vm$$-$$1
8.48 Vm$$-$$1
6.75 $$\times$$ 106 Vm$$-$$1
Explanation
H = 4.5 $$\times$$ 10$$-$$2
So B = $$\mu$$0$$\mu$$H
Thus $$E = {c \over n}B$$ (where n $$\Rightarrow$$ refractive index)
So $$E = {{3 \times {{10}^8} \times 4\pi \times {{10}^{ - 7}} \times 1.61 \times 4.5 \times {{10}^{ - 2}}} \over {\sqrt {1.61 \times 6.44} }}$$
$$E = 8.48$$
Comments (0)
