JEE MAIN - Physics (2021 - 25th February Morning Shift - No. 24)
The electric field in a region is given by $$\overrightarrow E = \left( {{3 \over 5}{E_0}\widehat i + {4 \over 5}{E_0}\widehat j} \right){N \over C}$$. The ratio of flux of reported field through the rectangular surface of area 0.2 m2 (parallel to y $$-$$ z plane) to that of the surface of area 0.3 m2 (parallel to x $$-$$ z plane) is a : b, where a = __________ [Here $${\widehat i}$$, $${\widehat j}$$ and $${\widehat k}$$ are unit vectors along x, y and z-axes respectively.]
Answer
1
Explanation
$$\phi = \overrightarrow E \,.\,\overrightarrow A $$
$${\overrightarrow A _a} = 0.2\widehat i$$
$${\overrightarrow A _b} = 0.3\widehat j$$
$${\phi _a} = \left( {{3 \over 5}{E_0}\widehat i + {4 \over 5}{E_0}\widehat j} \right).\,0.2\widehat i$$
$$ \Rightarrow $$ $${\phi _a} = {3 \over 5}{E_0} \times 0.2$$
$${\phi _b} = \left( {{3 \over 5}{E_0}\widehat i + {4 \over 5}{E_0}\widehat j} \right).\,0.3\widehat j$$
$$ \Rightarrow $$ $${\phi _b} = {4 \over 5}{E_0} \times 0.3$$
$${a \over b} = {{{\phi _a}} \over {{\phi _b}}} = {{{3 \over 5}{E_0} \times 0.2} \over {{4 \over 5}{E_0} \times 0.3}} = {6 \over {12}} = 0.5$$
$${\overrightarrow A _a} = 0.2\widehat i$$
$${\overrightarrow A _b} = 0.3\widehat j$$
$${\phi _a} = \left( {{3 \over 5}{E_0}\widehat i + {4 \over 5}{E_0}\widehat j} \right).\,0.2\widehat i$$
$$ \Rightarrow $$ $${\phi _a} = {3 \over 5}{E_0} \times 0.2$$
$${\phi _b} = \left( {{3 \over 5}{E_0}\widehat i + {4 \over 5}{E_0}\widehat j} \right).\,0.3\widehat j$$
$$ \Rightarrow $$ $${\phi _b} = {4 \over 5}{E_0} \times 0.3$$
$${a \over b} = {{{\phi _a}} \over {{\phi _b}}} = {{{3 \over 5}{E_0} \times 0.2} \over {{4 \over 5}{E_0} \times 0.3}} = {6 \over {12}} = 0.5$$
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