JEE MAIN - Physics (2013 (Offline) - No. 5)
The magnetic field in a travelling electromagnetic wave has a peak value of $$20$$ $$n$$$$T$$. The peak value of electric field strength is :
$$3V/m$$
$$6V/m$$
$$9V/m$$
$$12V/m$$
Explanation
From question,
$${B_0} = 20nT = 20 \times {10^{ - 9}}T$$
( as velocity of light in vacuum $$C = 3 \times {10^8}\,\,m{s^{ - 1}}$$ )
$${\overrightarrow E _0} = {\overrightarrow B _0} \times \overrightarrow C $$
$$\left| {{{\overrightarrow E }_0}} \right| = \left| {\overrightarrow B } \right|.\left| {\overrightarrow C } \right| = 20 \times {10^{ - 9}} \times 3 \times {10^8}$$
$$ = 6\,\,V/m.$$
$${B_0} = 20nT = 20 \times {10^{ - 9}}T$$
( as velocity of light in vacuum $$C = 3 \times {10^8}\,\,m{s^{ - 1}}$$ )
$${\overrightarrow E _0} = {\overrightarrow B _0} \times \overrightarrow C $$
$$\left| {{{\overrightarrow E }_0}} \right| = \left| {\overrightarrow B } \right|.\left| {\overrightarrow C } \right| = 20 \times {10^{ - 9}} \times 3 \times {10^8}$$
$$ = 6\,\,V/m.$$
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