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JEE Main 20199 AprilMorning Shift
Exam
JEE Main
Year
2019
Date
9 April
Shift
Morning Shift
Subject
Physics
Chapter
Electromagnetic Waves

JEE Main 2019 · 9 April · Morning Shift · Physics · Electromagnetic Waves

The magnetic field of a plane electromagnetic wave is given by → B...

The magnetic field of a plane electromagnetic wave is given by \overset{\to}{B} = {B}_{0}\overset{^}{i}\left[\cos \left(kz - \omegat\right)\right] + {B}_{1}\overset{^}{j}\cos \left(kz + \omegat\right), where B0=3×10–5T{B}_{0} = 3 \times {10}^{– 5}T and B1=2×10–6T{B}_{1} = 2 \times {10}^{– 6}T. The RMSRMS value of the force experienced by a stationary charge Q=10–4CQ = {10}^{– 4}C at z=0z = 0 is closest to

Options

  1. A0.1N0.1N
  2. B0.9N0.9N
  3. C3×10–2N3 \times {10}^{– 2}N
  4. D0.6N0.6N

Correct answer

D. 0.6N0.6N

Step-by-step solution

\overset{\to}{B} = {B}_{0}\overset{^}{i}\left[\cos \left(kz - \omegat\right)\right] + {B}_{1}\overset{^}{j}\cos \left(kz + \omegat\right)
B0=3×10−5T and B1=2×10−6T{B}_{0} = 3 \times {10}^{- 5}T \text{ and } {B}_{1} = 2 \times {10}^{- 6}T
Electric filed associated with it is,
\overset{\to}{E} = - {B}_{0}c\overset{^}{j}\cos \left(kz - \omegat\right) - {B}_{1}c\overset{^}{i}\cos \left(kz + \omegat\right)
Here, cc is the speed of light in vacuum.
At z=0,z = 0,
Frms=(QB0c)2+(QB1c)22{F}_{\text{rms}} = \sqrt{\dfrac{{\left(Q{B}_{0}c\right)}^{2} + {\left(Q{B}_{1}c\right)}^{2}}{2}}
Here cc is speed of light in vacuum.
=(10−4×3×10−5×3×108)2+(10−4×2×10−6×3×108)22= \sqrt{\dfrac{{\left({10}^{- 4} \times 3 \times {10}^{- 5} \times 3 \times {10}^{8}\right)}^{2} + {\left({10}^{- 4} \times 2 \times {10}^{- 6} \times 3 \times {10}^{8}\right)}^{2}}{2}}
=0.81+0.00362=0.6N= \sqrt{\dfrac{0.81 + 0.0036}{2}} = 0.6N
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