The electric field in an electromagnetic wave is given by . The magnetic field induction of this wave is (in SI unit):

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Q.
The electric field in an electromagnetic wave is given by $\vec{E} = \hat{i} 40 \cos ω (t - \frac{z}{c}) {NC}^{- 1}$ . The magnetic field induction of this wave is (in SI unit): [2024]

1 $\vec{B} = \hat{j} 40 \cos ω (t - \frac{z}{c})$

2 $\vec{B} = \hat{k} \frac{40}{c} \cos ω (t - \frac{z}{c})$

3 $\vec{B} = \hat{i} \frac{40}{c} \cos ω (t - \frac{z}{c})$

4 $\vec{B} = \hat{j} \frac{40}{c} \cos ω (t - \frac{z}{c})$

Ans.
(4)

$\vec{E} = 40 \cos (ω t - \frac{ω}{c} z) {NC}^{- 1} \hat{i}$

$\vec{E}$ is along $+ x$ direction

$\vec{v}$ is along $+ z$ direction

and Direction of $\hat{c}$ must be along $\vec{E} \times \vec{B}$ .

So direction of $\vec{B}$ will be along $+ y$ and magnitude of $\vec{B}$ will be $\frac{E_{0}}{c} .$

$B_{0} = \frac{E_{0}}{c} = \frac{40}{c}$

Solution

Q.
The electric field in an electromagnetic wave is given by $\vec{E} = \hat{i} 40 \cos ω (t - \frac{z}{c}) {NC}^{- 1}$ . The magnetic field induction of this wave is (in SI unit): [2024]

1 $\vec{B} = \hat{j} 40 \cos ω (t - \frac{z}{c})$

2 $\vec{B} = \hat{k} \frac{40}{c} \cos ω (t - \frac{z}{c})$

3 $\vec{B} = \hat{i} \frac{40}{c} \cos ω (t - \frac{z}{c})$

4 $\vec{B} = \hat{j} \frac{40}{c} \cos ω (t - \frac{z}{c})$

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Solution

Q. The electric field in an electromagnetic wave is given by E→=i^ 40cosω(t-zc) NC-1. The magnetic field induction of this wave is (in SI unit): [2024]

1 B→=j^ 40cosω(t-zc)

2 B→=k^40ccosω(t-zc)

3 B→=i^40ccosω(t-zc)

4 B→=j^40ccosω(t-zc)

Ans. (4) E→=40cos(ωt-ωcz) NC-1i^ E→ is along +x direction v→ is along +z direction and Direction of c^ must be along E→×B→. So direction of B→ will be along +y and magnitude of B→ will be E0c. B0=E0c=40c

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Q.
The electric field in an electromagnetic wave is given by $\vec{E} = \hat{i} 40 \cos ω (t - \frac{z}{c}) {NC}^{- 1}$ . The magnetic field induction of this wave is (in SI unit): [2024]

1 $\vec{B} = \hat{j} 40 \cos ω (t - \frac{z}{c})$

2 $\vec{B} = \hat{k} \frac{40}{c} \cos ω (t - \frac{z}{c})$

3 $\vec{B} = \hat{i} \frac{40}{c} \cos ω (t - \frac{z}{c})$

4 $\vec{B} = \hat{j} \frac{40}{c} \cos ω (t - \frac{z}{c})$