Verified Solution • 2025–26 Session moderate

QJEEM20Ph294f: JEE Main 2020 Physics - Class 12 JEEM20Ph294f Solution: An electron is constrained to move along the y-axis with a…

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
E→=30j^sin⁡(1.5×107t−5×10−2x)\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)

Multiple Choice Options

  • A 4.8 ×\times 10–19 N
  • B 2.4 ×\times 10–18 N
  • C 3.2 ×\times 10–18 N
  • D 1.6 ×\times 10–18 N

Step-by-Step Verified Solution

E→=30j^sin⁡(1.5×107t−5×10−2x)V/m\overrightarrow E = 30\widehat j\sin (1.5 \times {10^7}t - 5 \times {10^{ - 2}}x)V/m

V = 1.5×1025×10−2{{1.5 \times {{10}^2}} \over {5 \times {{10}^{ - 2}}}} = 3 ×\times 108 = C

⇒B=E/C=301.5×107×5×10−2\Rightarrow B = E/C = {{30} \over {1.5 \times {{10}^7}}} \times 5 \times {10^{ - 2}}

=10−7Tesla= {10^{ - 7}}Tesla

⇒Fmax=q(V→×B→)=∣qVB∣\Rightarrow {F_{max}} = q\left( {\overrightarrow V \times \overrightarrow B } \right) = \left| {qVB} \right|

=1.6×10−19×0.1×3×108×10−7= 1.6 \times {10^{ - 19}} \times 0.1 \times 3 \times {10^8} \times {10^{ - 7}}

=4.8×10−19N= 4.8 \times {10^{ - 19}}N

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