JEE Main 2026 — Jan 22, Shift 2 — Physics solution

JEE Main2026Jan 22, Shift 2Physics

A laser beam has intensity of $4.0 \times 1 0^{14}$ W/m $^{2}$ . The amplitude of magnetic field associated with beam is _____ T. (Take $ϵ_{0} = 8.85 \times 1 0^{- 12}$ C $^{2}$ /Nm $^{2}$ and $c = 3 \times 1 0^{8}$ m/s)

A
2.0
B
18.3
C
5.5
D
1.83

Solution

Intensity of an EM wave: $I = \frac{1}{2} ϵ_{0} E_{0}^{2} c$ , so $E_{0} = \frac{2 I}{ϵ _{0} c}$ .

And $B_{0} = \frac{E _{0}}{c} = \frac{1}{c} \frac{2 I}{ϵ _{0} c}$ .

Plugging in: $B_{0} = \frac{1}{3 \times 1 0 ^{8}} \frac{2 \times 4 \times 1 0 ^{14}}{8.85 \times 1 0 ^{- 12} \times 3 \times 1 0 ^{8}}$ .

The argument of the square root is $\frac{8 \times 1 0 ^{14}}{2.655 \times 1 0 ^{- 3}} \approx 3.01 \times 1 0^{17}$ , so $\cdot \approx 5.49 \times 1 0^{8}$ .

$B_{0} \approx \frac{5.49 \times 1 0 ^{8}}{3 \times 1 0 ^{8}} \approx 1.83$ T.

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