An electromagnetic wave is propagating in vacuum along -j^ direction. The magnetic field of the wave is given by B→=2×10-8cosπ×1015t+yc k^ T. The electric field E→ of this wave is (c= speed of light)

(B) E→=6cosπ×1015t+yc i^ V m-1

(A) E→=4cosπ×1015t+yc j^ V m-1

(B) E→=6cosπ×1015t+yc i^ V m-1

(C) E→=6cosπ×1015t-yc j^ V m-1

(D) E→=4cosπ×1015t-yc i^ V m-1

TS EAMCET · Physics · Electromagnetic Waves

An electromagnetic wave is propagating in vacuum along $- \hat{j}$ direction. The magnetic field of the wave is given by $\vec{B} = (2 \times 10^{- 8}) \cos [π \times 10^{15} (t + \frac{y}{c})] \hat{k} T$ . The electric field $\vec{E}$ of this wave is ( $c =$ speed of light)

A $\vec{E} = (4) \cos [π \times 10^{15} (t + \frac{y}{c})] \hat{j} V m^{- 1}$
B $\vec{E} = (6) \cos [π \times 10^{15} (t + \frac{y}{c})] \hat{i} V m^{- 1}$
C $\vec{E} = (6) \cos [π \times 10^{15} (t - \frac{y}{c})] \hat{j} V m^{- 1}$
D $\vec{E} = (4) \cos [π \times 10^{15} (t - \frac{y}{c})] \hat{i} V m^{- 1}$

Verified Solution

Answer & Solution

Correct Answer

(B) $\vec{E} = (6) \cos [π \times 10^{15} (t + \frac{y}{c})] \hat{i} V m^{- 1}$

Step-by-step Solution

Detailed explanation

As we know, c=E0B0⇒E0=B0c=2×10-83×108=6 V m-1 Now, direction of propagation of wave =E^×B^⇒-j^=E^×k^⇒E^=i^ Angular frequency will remain the same. Therefore electric field can be written as,…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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