NEET 2025 — Physics solution

NEET2025Physics

The electric field in a plane electromagnetic wave is given by

$E_{z} = 60 cos (5 x + 1.5 \times 1 0^{9} t)$ V/m.

Then the expression for the corresponding magnetic field is (here subscripts denote the direction of the field):

A
$B_{y} = 2 \times 1 0^{- 7} cos (5 x + 1.5 \times 1 0^{9} t) T$
B
$B_{x} = 2 \times 1 0^{- 7} cos (5 x + 1.5 \times 1 0^{9} t) T$
C
$B_{z} = 60 cos (5 x + 1.5 \times 1 0^{9} t) T$
D
$B_{y} = 60 sin (5 x + 1.5 \times 1 0^{9} t) T$

Solution

Wave speed: $v = ω / k = (1.5 \times 1 0^{9}) /5 = 3 \times 1 0^{8}$ m/s, i.e. light in vacuum.

Amplitude: $B_{0} = E_{0} / c = 60/ (3 \times 1 0^{8}) = 2 \times 1 0^{- 7}$ T.

Direction: the wave travels along $- \overset{x}{^}$ (since the argument is $k x + ω t$ ). With $E$ along $\overset{z}{^}$ and $\overset{v}{^} \times \hat{E}$ giving the direction of $B$ via $E \times B$ along $\overset{v}{^}$ , we get $B$ along $\overset{y}{^}$ .

$∴ B_{y} = 2 \times 1 0^{- 7} cos (5 x + 1.5 \times 1 0^{9} t) T$ .

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