JEE MAIN - Physics (2019 - 10th January Morning Slot - No. 3)
If the magnetic field of a plane electromagnetic wave is given by (the speed of light = 3 × 108 B = 100 × 10–6
sin $$\left[ {2\pi \times 2 \times {{10}^{15}}\left( {t - {x \over c}} \right)} \right]$$ then the maximum electric field associated with it is -
4.5 $$ \times $$ 104 N/C
4 $$ \times $$ 104 N/C
6 $$ \times $$ 104 N/C
3 $$ \times $$ 104 N/C
Explanation
E0 = B0 $$ \times $$ C
= 100 $$ \times $$ 10$$-$$6 $$ \times $$ 3 $$ \times $$ 108
= 3 $$ \times $$ 104 N/C
$$ \therefore $$ correct answer is 3 $$ \times $$ 104 N/C
= 100 $$ \times $$ 10$$-$$6 $$ \times $$ 3 $$ \times $$ 108
= 3 $$ \times $$ 104 N/C
$$ \therefore $$ correct answer is 3 $$ \times $$ 104 N/C
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