Verified Solution • 2025–26 Session moderate

QJEEM19Ph82b9: JEE Main 2019 Physics - Class 12 JEEM19Ph82b9 Solution: The magnetic field of a plane electromagnetic wave is given by…

The magnetic field of a plane electromagnetic wave is given by :
</mo></mrow><mover><mi>B</mi><moaccent="true">―</mo></mover><mo>=</mo><mrowdata−mjx−texclass="ORD"><msub><mi>B</mi><mn>0</mn></msub></mrow><mrowdata−mjx−texclass="ORD"><mover><mi>i</mi><mo></mo></mover></mrow><mrowdata−mjx−texclass="INNER"><modata−mjx−texclass="OPEN">[</mo><mrowdata−mjx−texclass="ORD"><mi>cos</mi><modata−mjx−texclass="NONE">⁡</mo><mostretchy="false">(</mo><mi>k</mi><mi>z</mi><mo>−</mo><mi>ω</mi><mi>t</mi><mostretchy="false">)</mo></mrow><modata−mjx−texclass="CLOSE">]</mo></mrow><mo>+</mo><mrowdata−mjx−texclass="ORD"><msub><mi>B</mi><mi>i</mi></msub></mrow><mrowdata−mjx−texclass="ORD"><mover><mi>j</mi><mo></mo></mover></mrow><mi>cos</mi><modata−mjx−texclass="NONE">⁡</mo><mostretchy="false">(</mo><mi>k</mi><mi>z</mi><mo>+</mo><mi>ω</mi><mi>t</mi><mostretchy="false">)</mo></math></mjx−assistive−mml></mjx−container></mjx−assistive−mml></mjx−container></mjx−assistive−mml></mjx−container></mo></mrow><mover><mi>B</mi><mo accent="true">―</mo></mover><mo>=</mo><mrow data-mjx-texclass="ORD"><msub><mi>B</mi><mn>0</mn></msub></mrow><mrow data-mjx-texclass="ORD"><mover><mi>i</mi><mo>^</mo></mover></mrow><mrow data-mjx-texclass="INNER"><mo data-mjx-texclass="OPEN">[</mo><mrow data-mjx-texclass="ORD"><mi>cos</mi><mo data-mjx-texclass="NONE">⁡</mo><mo stretchy="false">(</mo><mi>k</mi><mi>z</mi><mo>−</mo><mi>ω</mi><mi>t</mi><mo stretchy="false">)</mo></mrow><mo data-mjx-texclass="CLOSE">]</mo></mrow><mo>+</mo><mrow data-mjx-texclass="ORD"><msub><mi>B</mi><mi>i</mi></msub></mrow><mrow data-mjx-texclass="ORD"><mover><mi>j</mi><mo>^</mo></mover></mrow><mi>cos</mi><mo data-mjx-texclass="NONE">⁡</mo><mo stretchy="false">(</mo><mi>k</mi><mi>z</mi><mo>+</mo><mi>ω</mi><mi>t</mi><mo stretchy="false">)</mo></math></mjx-assistive-mml></mjx-container></mjx-assistive-mml></mjx-container></mjx-assistive-mml></mjx-container> B0 = 3 × 10–5 T and B1 = 2 × 10–6 T.
The rms value of the force experienced by a stationary charge Q = 10–4 C at z = 0 is closest to :

Multiple Choice Options

  • A 0.6 N
  • B 0.9 N
  • C 3 × 10–2 N
  • D 0.1 N

Step-by-Step Verified Solution

Maximum electric field E = (B) (C)

E0→=(3×10−5)c(−j^)\overrightarrow {{E_0}} = \left( {3 \times {{10}^{ - 5}}} \right)c\left( { - \widehat j} \right)

E1→=(2×10−6)c(−i^)\overrightarrow {{E_1}} = \left( {2 \times {{10}^{ - 6}}} \right)c\left( { - \widehat i} \right)

Maximum force

F→net=10−4×3×108(3×10−5)2+(2×10−6)2=0.9N{\overrightarrow F _{net}} = {10^{ - 4}} \times 3 \times {10^8}\sqrt {{{\left( {3 \times {{10}^{ - 5}}} \right)}^2} + {{\left( {2 \times {{10}^{ - 6}}} \right)}^2}} = 0.9N

Frms=F02=0.6 N{F_{rms}} = {{{F_0}} \over {\sqrt 2 }} = 0.6\,N (approx)

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