JEE MAIN - Physics (2020 - 5th September Morning Slot - No. 11)
An electron is constrained to move along
the y-axis with a speed of 0.1 c (c is the
speed of light) in the presence of
electromagnetic wave, whose electric
field is
$$\overrightarrow E = 30\widehat j\sin \left( {1.5 \times {{10}^7}t - 5 \times {{10}^{ - 2}}x} \right)$$ V/m.
The maximum magnetic force experienced by the electron will be :
(given c = 3 $$ \times $$ 108 ms–1 and electron charge = 1.6 $$ \times $$ 10–19 C)
$$\overrightarrow E = 30\widehat j\sin \left( {1.5 \times {{10}^7}t - 5 \times {{10}^{ - 2}}x} \right)$$ V/m.
The maximum magnetic force experienced by the electron will be :
(given c = 3 $$ \times $$ 108 ms–1 and electron charge = 1.6 $$ \times $$ 10–19 C)
4.8 $$ \times $$ 10–19 N
2.4 $$ \times $$ 10–18 N
3.2 $$ \times $$ 10–18 N
1.6 $$ \times $$ 10–18 N
Explanation
$$
\overrightarrow E = 30\widehat j\sin (1.5 \times {10^7}t - 5 \times {10^{ - 2}}x)V/m$$
V = $${{1.5 \times {{10}^2}} \over {5 \times {{10}^{ - 2}}}}$$ = 3 $$ \times $$ 108 = C
$$ \Rightarrow B = E/C = {{30} \over {1.5 \times {{10}^7}}} \times 5 \times {10^{ - 2}}$$
$$ = {10^{ - 7}}Tesla$$
$$ \Rightarrow {F_{max}} = q\left( {\overrightarrow V \times \overrightarrow B } \right) = \left| {qVB} \right|$$
$$ = 1.6 \times {10^{ - 19}} \times 0.1 \times 3 \times {10^8} \times {10^{ - 7}}$$
$$ = 4.8 \times {10^{ - 19}}N$$
V = $${{1.5 \times {{10}^2}} \over {5 \times {{10}^{ - 2}}}}$$ = 3 $$ \times $$ 108 = C
$$ \Rightarrow B = E/C = {{30} \over {1.5 \times {{10}^7}}} \times 5 \times {10^{ - 2}}$$
$$ = {10^{ - 7}}Tesla$$
$$ \Rightarrow {F_{max}} = q\left( {\overrightarrow V \times \overrightarrow B } \right) = \left| {qVB} \right|$$
$$ = 1.6 \times {10^{ - 19}} \times 0.1 \times 3 \times {10^8} \times {10^{ - 7}}$$
$$ = 4.8 \times {10^{ - 19}}N$$
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