Question

JEE Mains physics question is in the image

Answer 1

ANSWER:- Option(A)

SOLUTION:-

$\implies E_{o} = B_{o} C$

$\implies B_{o}= \frac{\in_{o}}{c} \\ \\ \\ \implies B_{o}=\in_{o} \sqrt{ \mu_{o}\in_{o}}$

NOW,

$\implies B= B_{o} \cos(kx) \hat{k} \\ \\ \\ \implies \: \boxed{ B =E_{o} \sqrt{ \mu _{o} \in_{o} } \cos(kx \: ) \hat{k}}$

HENCE, OPTION (A) Is the answer.

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HOPE IT HELPS :)

Answer 2

Answer:

According to the definition of an EM wave,

→ E × B = EM Wave's magnitude along with direction.

where, E × B refers to the cross product of Electric Field vector and Magnetic Field Vector.

According to the question, Electric field is in the direction of j. Assuming EM Wave's direction to be i, we get:

j × x = i

⇒The direction 'x' should be k.

Hence we can eliminate option C and D.

Now coming to the equation, we are given that t = 0. Hence we get the phase difference to be kx. According to the definition, we also get E/B = c

Therefore, we can write B as:

→ B = E/c

→ B = E₀.Cos(kx)/c (k)

'c' can also be expressed as 1/√μ₀ε₀.

Hence we get:

→ B = E₀.Cos(kx)/ (1/√μ₀ε₀)    in k direction.

→ B = E₀.Cos(kx) (√μ₀ε₀)   in k direction.

Hence Option A is the correct answer.