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The electric field in a plane electromagnetic wave is given by \({\mathrm{E}}_{\mathrm{z}}=60\cos \left(5\mathrm{x}+1.5\…

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The electric field in a plane electromagnetic wave is given by \({\mathrm{E}}_{\mathrm{z}}=60\cos \left(5\mathrm{x}+1.5\times {10}^{9}\mathrm{t}\right)\mathrm{V}/\mathrm{m}.\)
Then expression for the coresponding magnetic field is (here subscripts denote the direction of the field) :

[NEET 2025]

a

\({\mathrm{B}}_{\mathrm{z}}=60\cos \left(5\mathrm{x}+1.5\times {10}^{9}\mathrm{t}\right)\mathrm{T}\)

b

\({\mathrm{B}}_{\mathrm{y}}=60\sin \left(5\mathrm{x}+1.5\times {10}^{9}\mathrm{t}\right)\mathrm{T}\)

c

\({\mathrm{B}}_{\mathrm{y}}=2\times {10}^{-7}\cos \left(5\mathrm{x}+1.5\times {10}^{9}t\right)\mathrm{T}\)

d

\({\mathrm{B}}_{\mathrm{x}}=2\times {10}^{-7}\cos \left(5\mathrm{x}+1.5\times {10}^{9}\mathrm{t}\right)\mathrm{T}\)

✓ Correct answer: c)

\({\mathrm{B}}_{\mathrm{y}}=2\times {10}^{-7}\cos \left(5\mathrm{x}+1.5\times {10}^{9}t\right)\mathrm{T}\)

Explanation

In electromagnetic wave, E and B are in same phase and \(B_0=\frac{E_0}{c}\); their planes are perpendicular to each other.

\(\begin{array}{l}\therefore B_y=\frac{60}{c} \cos \left(5 x+1.5 \times 10^9 t\right) T \\ =\frac{60}{3 \times 10^8} \cos \left(5 x+1.5 \times 10^9 t\right) T \\ B_y=2 \times 10^{-7} \cos \left(5 x+1.5 \times 10^9 t\right) T \end{array}\)

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