JEE MAIN - Physics (2017 - 8th April Morning Slot - No. 4)
Magnetic field in a plane electromagnetic wave is given by
$$\overrightarrow B $$ = B0 sin (k x + $$\omega $$t) $$\widehat j\,T$$
Expression for corresponding electric field will be :
Where c is speed of light.
$$\overrightarrow B $$ = B0 sin (k x + $$\omega $$t) $$\widehat j\,T$$
Expression for corresponding electric field will be :
Where c is speed of light.
$$\overrightarrow E $$ = B0 c sin (k x + $$\omega $$t) $$\widehat k$$ V/m
$$\overrightarrow E $$ = $${{{B_0}} \over c}$$ sin (k x + $$\omega $$t) $$\widehat k$$ V/m
$$\overrightarrow E $$ = $$-$$ B0 c sin (kx +$$\omega $$t) $$\widehat k$$ V/m
$$\overrightarrow E $$ = B0 c sin (kx $$-$$$$\omega $$t) $$\widehat k$$ V/m
Explanation
The relation between electric and magnetic field is ,
C = $${{\overrightarrow E } \over {\overrightarrow B }}$$
$$ \Rightarrow $$$$\,\,\,$$ $$\overrightarrow E $$ = C $$\overrightarrow B $$
Electric field component is perpendicular to the direction of magnetic field. Given magnetic field is along y $$-$$ axis,
So, electric field along z $$-$$ axis will be
$$\overrightarrow E $$ = B0 C sin (kx + $$\omega $$t) $$\,\widehat k$$ v/m
C = $${{\overrightarrow E } \over {\overrightarrow B }}$$
$$ \Rightarrow $$$$\,\,\,$$ $$\overrightarrow E $$ = C $$\overrightarrow B $$
Electric field component is perpendicular to the direction of magnetic field. Given magnetic field is along y $$-$$ axis,
So, electric field along z $$-$$ axis will be
$$\overrightarrow E $$ = B0 C sin (kx + $$\omega $$t) $$\,\widehat k$$ v/m
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