JEE MAIN - Physics (2020 - 7th January Morning Slot - No. 4)
If the magnetic field in a plane electromagnetic wave is given by
$$\overrightarrow B $$ = 3 $$ \times $$ 10-8 sin(1.6 $$ \times $$ 103x + 48 $$ \times $$ 1010t)$$\widehat j$$ T, then what will be expression for electric field ?
$$\overrightarrow B $$ = 3 $$ \times $$ 10-8 sin(1.6 $$ \times $$ 103x + 48 $$ \times $$ 1010t)$$\widehat j$$ T, then what will be expression for electric field ?
$$\overrightarrow E $$ = (9sin(1.6 $$ \times $$ 103x + 48 $$ \times $$ 1010t)$$\widehat k$$ V/m)
$$\overrightarrow E $$ = (60sin(1.6 $$ \times $$ 103x + 48 $$ \times $$ 1010t)$$\widehat k$$ V/m)
$$\overrightarrow E $$ = (3 $$ \times $$ 10-8 sin(1.6 $$ \times $$ 103x + 48 $$ \times $$ 1010t)$$\widehat i$$ V/m)
$$\overrightarrow E $$ = (3 $$ \times $$ 10-8 sin(1.6 $$ \times $$ 103x + 48 $$ \times $$ 1010t)$$\widehat j$$ V/m)
Explanation
Given $$\overrightarrow B $$ = 3 $$ \times $$ 10-8 sin(1.6 $$ \times $$ 103x + 48 $$ \times $$ 1010t)$$\widehat j$$ T
We know, $${{{E_0}} \over {{B_0}}} = c$$
$$ \Rightarrow $$ E0 = (3 $$ \times $$ 10-8) $$ \times $$ (3 $$ \times $$ 10-8) = 9 V/m
$$ \therefore $$ $$\overrightarrow E $$ = (9sin(1.6 $$ \times $$ 103x + 48 $$ \times $$ 1010t)$$\widehat k$$ V/m)
We know, $${{{E_0}} \over {{B_0}}} = c$$
$$ \Rightarrow $$ E0 = (3 $$ \times $$ 10-8) $$ \times $$ (3 $$ \times $$ 10-8) = 9 V/m
$$ \therefore $$ $$\overrightarrow E $$ = (9sin(1.6 $$ \times $$ 103x + 48 $$ \times $$ 1010t)$$\widehat k$$ V/m)
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