The magnetic field of an E.M. wave is given by: B = ( 32 i + 12 j) 30 ( ( t - zc)) The corresponding electric field in S.I. units is:

The magnetic field of an E.M. wave is given by: B = ( 32 i + 12 j… — Solution | JEE Physics

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The magnetic field of an E.M. wave is given by: $\vec{B} = \left( \frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j} \right) 30 \sin \left( \omega \left( t - \frac{z}{c} \right) \right)$ The corresponding electric field in S.I. units is:

A
$\vec{E} = \left( \frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j} \right) 30 c \sin \left( \omega \left( t + \frac{z}{c} \right) \right)$
B
$\vec{E} = \left( \frac{3}{4} \hat{i} + \frac{1}{4} \hat{j} \right) 30 c \cos \left( \omega \left( t - \frac{z}{c} \right) \right)$
C
$\vec{E} = \left( \frac{\sqrt{3}}{2} \hat{i} - \frac{1}{2} \hat{j} \right) 30 c \sin \left( \omega \left( t + \frac{z}{c} \right) \right)$
D
$\vec{E} = \left( \frac{1}{2} \hat{i} - \frac{\sqrt{3}}{2} \hat{j} \right) 30 c \sin \left( \omega \left( t - \frac{z}{c} \right) \right)$

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: The magnetic field is

\vec{B} = \left( \frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j} \right) 30 \sin \left[ \omega \left( t - \frac{z}{c} \right) \right]

Find: The corresponding electric field $\vec{E}$ in S.I. units.

For an electromagnetic wave, the electric and magnetic fields are perpendicular and related through the speed of light. The phase factor $\omega\left(t-\frac{z}{c}\right)$ shows that the wave propagates along the positive $z$ -direction.

The solution uses the relation

\vec{E} = \vec{B} \times \vec{c}, \qquad E_0 = B_0 c

so the electric field has magnitude

E_0 = 30c

Direction and final expression

Using the working shown in the solution, the final direction of the electric field is obtained as

\vec{E} = \left( \frac{1}{2} \hat{i} - \frac{\sqrt{3}}{2} \hat{j} \right) 30c \sin \left[ \omega \left( t - \frac{z}{c} \right) \right]

Therefore, the correct option is D.

Common mistakes

Assuming $\vec{E}$ is parallel to $\vec{B}$ . In an electromagnetic wave, $\vec{E}$ and $\vec{B}$ are mutually perpendicular. Always use the propagation direction together with the perpendicularity condition.
Missing the sign of propagation from the phase term. The expression $\omega\left(t-\frac{z}{c}\right)$ represents propagation along positive $z$ , so the direction of $\vec{E}$ must be chosen consistently with the right-hand rule.
Using the wrong magnitude relation between the fields. In S.I. units, the amplitudes satisfy $E_0 = cB_0$ , not $E_0 = \frac{B_0}{c}$ . This changes the coefficient in the final option.

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Answer

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