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Question: The electric field component of a monochromatic radiation is given by \[\overrightarrow E = 2{E_0}_{...

The electric field component of a monochromatic radiation is given by E→=2E0i^cos⁡kzcos⁡ωt\overrightarrow E = 2{E_0}_{}\hat i\cos kz\cos \omega t. Its magnetic field B→\overrightarrow B is given by :
(A) 2E0cj^sin⁡kzcos⁡ωt{\dfrac{{2{E_0}}}{c}_{}}\hat j\sin kz\cos \omega t
(B) −2E0cj^sin⁡kzsin⁡ωt - {\dfrac{{2{E_0}}}{c}_{}}\hat j\sin kz\sin \omega t
(C) 2E0cj^sin⁡kzsin⁡ωt{\dfrac{{2{E_0}}}{c}_{}}\hat j\sin kz\sin \omega t
(D) 2E0cj^cos⁡kzcos⁡ωt{\dfrac{{2{E_0}}}{c}_{}}\hat j\cos kz\cos \omega t

Explanation

Solution

Hint Apply faraday’s law of induction for the given electric field. According to faraday’s law of induction , dEdz=−dBdt\dfrac{{dE}}{{dz}} = - \dfrac{{dB}}{{dt}}. Use this formula and find out the B vector value.

Complete Step By Step Solution
For the given electric field of monochromatic radiation, we apply faraday’s law of induction since we need to find the magnetic field intensity B.
Now according to faraday’s law of induction,
dEdz=−dBdt\dfrac{{dE}}{{dz}} = - \dfrac{{dB}}{{dt}}
Which means,
dEdz=−dEdt\dfrac{{dE}}{{dz}} = - \dfrac{{dE}}{{dt}}
dEdz=−d(2E0j^cos⁡kzcos⁡ωt)dt\dfrac{{dE}}{{dz}} = - \dfrac{{d(2{E_0}\hat j\cos kz\cos \omega t)}}{{dt}}
Differentiating the above equation with respect to z we get,
dEdz=(2E0j^ksin⁡kzcos⁡ωt)=−dBdt\dfrac{{dE}}{{dz}} = (2{E_0}\hat jk\sin kz\cos \omega t) = - \dfrac{{dB}}{{dt}}
(Differentiation of cos⁡kz\cos kzis sin⁡kz\sin kzmultiplied by the differentiation of kz )
Now taking dtdtto the other side, we get
⇒−(2E0j^ksin⁡kzcos⁡ωt)×dt=−dB\Rightarrow - (2{E_0}{\hat j_{}}k\sin kz\cos \omega t) \times dt = - dB
To remove the differentiation term, we need to integrate on both sides. Now integration of dBdB will give us B value. Let’s do the integration on the left hand side.
⇒B=∫(2E0j^ksin⁡kzcos⁡ωt)×dt\Rightarrow B = \int {(2{E_0}{{\hat j}_{}}k\sin kz\cos \omega t) \times dt}
Except the cos⁡ωt\cos \omega tterm, rest all term remains the same as constants. Now integration of cos⁡ωt\cos \omega tis equal to sin⁡ωt\sin \omega tmultiplied by integration of ωt\omega t.
⇒B=−2E0j^ksin⁡kzsin⁡ωt×1ω\Rightarrow B = - 2{E_0}{\hat j_{}}k\sin kz\sin \omega t \times \dfrac{1}{\omega }(integration of ωt\omega tis 1ω\dfrac{1}{\omega })
Now the amplitudes of the waves are related using the equation,
E0B0=ωk=C\dfrac{{{E_0}}}{{{B_0}}} = \dfrac{\omega }{k} = C
Substituting the value of ωk\dfrac{\omega }{k}in the equation for B we get
⇒B=−2E0j^sin⁡kzsin⁡ωt×kω\Rightarrow B = - 2{E_0}\hat j\sin kz\sin \omega t \times \dfrac{k}{\omega }
⇒B=−2E0Cj^sin⁡kzsin⁡ωt\Rightarrow B = \dfrac{{ - 2{E_0}}}{C}\hat j\sin kz\sin \omega t

Hence, Option (c) is the right answer for the given question.

Note Faraday’s law of induction states that there will be a current induced on the body or conductor when it is exposed to a change in the magnetic field. This law led to the birth of alternating current and the concept of EMF. Faraday used the bar magnet, and brought it near a closed-loop and observed the changes in the current value. This led him to ideate that change in magnetic flux with respect to time induced an EMF on the coil.