JEE MAIN - Physics (2021 - 16th March Morning Shift - No. 10)
A plane electromagnetic wave of frequency 500 MHz is travelling in vacuum along y-direction. At a particular point in space and time,
$$\overrightarrow B $$ = 8.0 $$\times$$ 10$$-$$8 $$\widehat z$$T. The value of electric field at this point is :
(speed of light = 3 $$\times$$ 108 ms$$-$$1)
$$\widehat x$$, $$\widehat y$$, $$\widehat z$$ are unit vectors along x, y and z directions.
$$\overrightarrow B $$ = 8.0 $$\times$$ 10$$-$$8 $$\widehat z$$T. The value of electric field at this point is :
(speed of light = 3 $$\times$$ 108 ms$$-$$1)
$$\widehat x$$, $$\widehat y$$, $$\widehat z$$ are unit vectors along x, y and z directions.
2.6 $$\widehat x$$ V/m
$$-$$24 $$\widehat x$$ V/m
24 $$\widehat x$$ V/m
$$-$$2.6 $$\widehat y$$ V/m
Explanation
$${E_0} = B.C$$
$${E_0} = (8 \times {10^{ - 8}}) \times (3 \times {10^8})$$
$$ \Rightarrow {E_0} = 24$$
Direction of wave travelling is in $$\overrightarrow E \times \overrightarrow B $$
So, $$( - \widehat x) \times \widehat z = + \widehat y$$
$$ \therefore $$ $$\widehat E = -24\widehat x$$ V/m
$${E_0} = (8 \times {10^{ - 8}}) \times (3 \times {10^8})$$
$$ \Rightarrow {E_0} = 24$$
Direction of wave travelling is in $$\overrightarrow E \times \overrightarrow B $$
So, $$( - \widehat x) \times \widehat z = + \widehat y$$
$$ \therefore $$ $$\widehat E = -24\widehat x$$ V/m
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