JEE MAIN - Physics (2020 - 6th September Evening Slot - No. 17)
For a plane electromagnetic wave, the magnetic field at a point x and time t is
$$\overrightarrow B \left( {x,t} \right)$$ = $$\left[ {1.2 \times {{10}^{ - 7}}\sin \left( {0.5 \times {{10}^3}x + 1.5 \times {{10}^{11}}t} \right)\widehat k} \right]$$ T
The instantaneous electric field $$\overrightarrow E $$ corresponding to $$\overrightarrow B $$ is :
(speed of light c = 3 × 108 ms–1)
$$\overrightarrow B \left( {x,t} \right)$$ = $$\left[ {1.2 \times {{10}^{ - 7}}\sin \left( {0.5 \times {{10}^3}x + 1.5 \times {{10}^{11}}t} \right)\widehat k} \right]$$ T
The instantaneous electric field $$\overrightarrow E $$ corresponding to $$\overrightarrow B $$ is :
(speed of light c = 3 × 108 ms–1)
$$\overrightarrow E \left( {x,t} \right) = \left[ {36\sin \left( {1 \times {{10}^3}x + 1.5 \times {{10}^{11}}t} \right)\widehat i} \right]$$ $${V \over m}$$
$$\overrightarrow E \left( {x,t} \right) = \left[ {36\sin \left( {0.5 \times {{10}^3}x + 1.5 \times {{10}^{11}}t} \right)\widehat k} \right]{V \over m}$$
$$\overrightarrow E \left( {x,t} \right) = \left[ {36\sin \left( {1 \times {{10}^3}x + 0.5 \times {{10}^{11}}t} \right)\widehat j} \right]{V \over m}$$
$$\overrightarrow E \left( {x,t} \right) = \left[ { - 36\sin \left( {0.5 \times {{10}^3}x + 1.5 \times {{10}^{11}}t} \right)\widehat j} \right]{V \over m}$$
Explanation
Given,
$$\overrightarrow B \left( {x,t} \right)$$ = $$\left[ {1.2 \times {{10}^{ - 7}}\sin \left( {0.5 \times {{10}^3}x + 1.5 \times {{10}^{11}}t} \right)\widehat k} \right]$$ T
Wave is travelling along (–x) axis and $$\overrightarrow B $$ is along +z axis.
We know, Magnitude of electric field
E = BC
= 1.2 $$ \times $$ 10-7 sin ( 0.5 10 + 1.5$$ \times $$1011t ) $$ \times $$ 3 $$ \times $$ 108
= 36 sin (0.5$$ \times $$103x+1.5$$ \times $$1011t) V/m
Also, $$\overrightarrow s = {{\overrightarrow E \times \overrightarrow B } \over {{\mu _0}}}$$
$$ \Rightarrow $$ $$ - \widehat i = {{\overrightarrow E \times \widehat k} \over {{\mu _0}}}$$
$$ \therefore $$ Direction of $${\overrightarrow E = - \widehat j}$$
$$ \therefore $$ Instantaneous electric field,
$$\overrightarrow E \left( {x,t} \right) = \left[ { - 36\sin \left( {0.5 \times {{10}^3}x + 1.5 \times {{10}^{11}}t} \right)\widehat j} \right]{V \over m}$$
$$\overrightarrow B \left( {x,t} \right)$$ = $$\left[ {1.2 \times {{10}^{ - 7}}\sin \left( {0.5 \times {{10}^3}x + 1.5 \times {{10}^{11}}t} \right)\widehat k} \right]$$ T
Wave is travelling along (–x) axis and $$\overrightarrow B $$ is along +z axis.
We know, Magnitude of electric field
E = BC
= 1.2 $$ \times $$ 10-7 sin ( 0.5 10 + 1.5$$ \times $$1011t ) $$ \times $$ 3 $$ \times $$ 108
= 36 sin (0.5$$ \times $$103x+1.5$$ \times $$1011t) V/m
Also, $$\overrightarrow s = {{\overrightarrow E \times \overrightarrow B } \over {{\mu _0}}}$$
$$ \Rightarrow $$ $$ - \widehat i = {{\overrightarrow E \times \widehat k} \over {{\mu _0}}}$$
$$ \therefore $$ Direction of $${\overrightarrow E = - \widehat j}$$
$$ \therefore $$ Instantaneous electric field,
$$\overrightarrow E \left( {x,t} \right) = \left[ { - 36\sin \left( {0.5 \times {{10}^3}x + 1.5 \times {{10}^{11}}t} \right)\widehat j} \right]{V \over m}$$
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