JEE MAIN - Physics (2021 - 17th March Evening Shift - No. 23)
The electric field in a region is given by $$\overrightarrow E = {2 \over 5}{E_0}\widehat i + {3 \over 5}{E_0}\widehat j$$ with $${E_0} = 4.0 \times {10^3}{N \over C}$$. The flux of this field through a rectangular surface area 0.4 m2 parallel to the Y-Z plane is __________ Nm2C$$-$$1.
Answer
640
Explanation
$$\phi = \overrightarrow E \,.\,\overrightarrow A $$
$$ = {{{E_0}} \over 5}\left( {2\widehat i + 3\widehat j} \right)\,.\,\left( {0.4\widehat i} \right)$$
$$ = {{4000} \over 5}\left( {2 \times 0.4} \right)$$
$$ = 640$$ Nm2 C$$-$$1
$$ = {{{E_0}} \over 5}\left( {2\widehat i + 3\widehat j} \right)\,.\,\left( {0.4\widehat i} \right)$$
$$ = {{4000} \over 5}\left( {2 \times 0.4} \right)$$
$$ = 640$$ Nm2 C$$-$$1
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