JEE MAIN - Physics (2019 - 9th January Morning Slot - No. 8)
A plane electromagnetic wave of frequency 50 MHz travels in free space along the positive x-direction. At a particular point in space and time, $$\overrightarrow E = 6.3\widehat j\,V/m.$$ The corresponding magnetic field $$\overrightarrow {B,} $$ at that point will be :
18.9 $$ \times $$ 10$$-$$8 $$\widehat k$$T
2.1 $$ \times $$ 10$$-$$8 $$\widehat k$$T
6.3 $$ \times $$ 10$$-$$8 $$\widehat k$$T
18.9 $$ \times $$ 108 $$\widehat k$$T
Explanation
Given, $$\overrightarrow E = 6.3\widehat j$$
$$ \therefore $$ $$\left| {\overrightarrow E } \right| = \sqrt {{{\left( {6.3} \right)}^2}} = 6.3$$
$$ \therefore $$ $$\left| {\overrightarrow B } \right| = {{\left| {\overrightarrow E } \right|} \over C}$$
$$ = {{6.3} \over {3 \times {{10}^8}}}$$
$$ = 2.1 \times {10^{ - 8}}\,$$ T
Now we have to find the direction of $$\overrightarrow B $$.
We know, $$\widehat E \times \widehat B = \widehat C$$ and given $$\overrightarrow E $$ is in y-direction and wave moving in positive x-direction.
$$ \therefore $$ $$\widehat J \times \widehat B = \widehat i$$
$$ \Rightarrow $$ $$\widehat B = \widehat K$$
$$ \therefore $$ $$\overrightarrow B = \left| {\overrightarrow B } \right|\widehat B$$
$$ = 2.1 \times {10^{ - 8}}\,\widehat K\,T$$
$$ \therefore $$ $$\left| {\overrightarrow E } \right| = \sqrt {{{\left( {6.3} \right)}^2}} = 6.3$$
$$ \therefore $$ $$\left| {\overrightarrow B } \right| = {{\left| {\overrightarrow E } \right|} \over C}$$
$$ = {{6.3} \over {3 \times {{10}^8}}}$$
$$ = 2.1 \times {10^{ - 8}}\,$$ T
Now we have to find the direction of $$\overrightarrow B $$.
We know, $$\widehat E \times \widehat B = \widehat C$$ and given $$\overrightarrow E $$ is in y-direction and wave moving in positive x-direction.
$$ \therefore $$ $$\widehat J \times \widehat B = \widehat i$$
$$ \Rightarrow $$ $$\widehat B = \widehat K$$
$$ \therefore $$ $$\overrightarrow B = \left| {\overrightarrow B } \right|\widehat B$$
$$ = 2.1 \times {10^{ - 8}}\,\widehat K\,T$$
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