The magnetic field of an E.M. wave is given by (SI Units) The corresponding electric field in S.I. units is: [2025]

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Q.
The magnetic field of an E.M. wave is given by

$\vec{B} = (\frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j}) 30 \sin [ω (t - \frac{z}{c})]$ (SI Units)

The corresponding electric field in S.I. units is: [2025]

1 $\vec{E} = (\frac{1}{2} \hat{i} - \frac{\sqrt{3}}{2} \hat{j}) 30 c \sin [ω (t - \frac{z}{c})]$

2 $\vec{E} = (\frac{3}{4} \hat{i} + \frac{1}{4} \hat{j}) 30 c \cos [ω (t - \frac{z}{c})]$

3 $\vec{E} = (\frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j}) 30 c \sin [ω (t + \frac{z}{c})]$

4 $\vec{E} = (\frac{\sqrt{3}}{2} \hat{i} - \frac{1}{2} \hat{j}) 30 c \sin [ω (t + \frac{z}{c})]$

Ans.
(1)

$\vec{B} = (\frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j}) 30 \sin [ω (t - \frac{z}{c})]$

$\vec{E} \times \vec{B} = \vec{c} and E_{0} = B_{0} c$

Here $E = (\frac{\sqrt{3}}{2} (- \hat{j}) + \frac{1}{2} \hat{i})$

$E_{0} = 30 c$

$\vec{E} = (\frac{1}{2} \hat{i} - \frac{\sqrt{3}}{2} \hat{j}) 30 c \sin [ω (t - \frac{z}{c})]$

Solution

Q.
The magnetic field of an E.M. wave is given by

$\vec{B} = (\frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j}) 30 \sin [ω (t - \frac{z}{c})]$ (SI Units)

The corresponding electric field in S.I. units is: [2025]

1 $\vec{E} = (\frac{1}{2} \hat{i} - \frac{\sqrt{3}}{2} \hat{j}) 30 c \sin [ω (t - \frac{z}{c})]$

2 $\vec{E} = (\frac{3}{4} \hat{i} + \frac{1}{4} \hat{j}) 30 c \cos [ω (t - \frac{z}{c})]$

3 $\vec{E} = (\frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j}) 30 c \sin [ω (t + \frac{z}{c})]$

4 $\vec{E} = (\frac{\sqrt{3}}{2} \hat{i} - \frac{1}{2} \hat{j}) 30 c \sin [ω (t + \frac{z}{c})]$

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Solution

Q. The magnetic field of an E.M. wave is given by B→=(32i^+12j^)30sin[ω(t–zc)] (SI Units) The corresponding electric field in S.I. units is: [2025]

1 E→=(12i^–32j^)30c sin[ω(t–zc)]

2 E→=(34i^+14j^)30c cos[ω(t–zc)]

3 E→=(12i^+32j^)30c sin[ω(t+zc)]

4 E→=(32i^–12j^)30c sin[ω(t+zc)]

Ans. (1) B→=(32i^+12j^)30sin[ω(t–zc)] E→×B→=c→ and E0=B0c Here E=(32(–j^)+12i^) E0=30c E→=(12i^–32j^) 30c sin [ω(t–zc)]

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Q.
The magnetic field of an E.M. wave is given by

$\vec{B} = (\frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j}) 30 \sin [ω (t - \frac{z}{c})]$ (SI Units)

The corresponding electric field in S.I. units is: [2025]

1 $\vec{E} = (\frac{1}{2} \hat{i} - \frac{\sqrt{3}}{2} \hat{j}) 30 c \sin [ω (t - \frac{z}{c})]$

2 $\vec{E} = (\frac{3}{4} \hat{i} + \frac{1}{4} \hat{j}) 30 c \cos [ω (t - \frac{z}{c})]$

3 $\vec{E} = (\frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j}) 30 c \sin [ω (t + \frac{z}{c})]$

4 $\vec{E} = (\frac{\sqrt{3}}{2} \hat{i} - \frac{1}{2} \hat{j}) 30 c \sin [ω (t + \frac{z}{c})]$