Modern Physics 7 Question 82

12. The magnetic field of an electromagnetic wave is given by $\mathbf{B}=1.6 \times 10^{-6} \cos \left(2 \times 10^{7} z+6 \times 10^{15} t\right)(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}) Wbm^{-2}$

The associated electric field will be (Main 2019, 8 April II)

(a) $\mathbf{E}=4.8 \times 10^{2} \cos \left(2 \times 10^{7} z-6 \times 10^{15} t\right)(-2 \hat{\mathbf{j}}+\hat{\mathbf{i}}) Vm^{-1}$

(b) $\mathbf{E}=4.8 \times 10^{2} \cos \left(2 \times 10^{7} z-6 \times 10^{15} t\right)(2 \hat{\mathbf{j}}+\hat{\mathbf{i}}) Vm^{-1}$

(c) $\mathbf{E}=4.8 \times 10^{2} \cos \left(2 \times 10^{7} z+6 \times 10^{15} t\right)(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}) Vm^{-1}$

(d) $\mathbf{E}=4.8 \times 10^{2} \cos \left(2 \times 10^{7} z+6 \times 10^{15} t\right)(-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}) Vm^{-1}$

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Solution:

  1. Given,

$\mathbf{B}=1.6 \times 10^{-6} \cos \left(2 \times 10^{7} z+6 \times 10^{15} t\right)(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}) Wbm^{-2}$

From the given equation, it can be said that the electromagnetic wave is propagating negative $z$-direction, i.e. $-\hat{\mathbf{k}}$

Equation of associated electric field will be

$$ \mathbf{E}=(|\mathbf{B}| c) \cos (k z+\omega t) \cdot \hat{\mathbf{n}} $$

where, $\hat{\mathbf{n}}=$ a vector perpendicular to $\mathbf{B}$.

So, $|\mathbf{E}|=|\mathbf{B}| \cdot c$

$$ =1.6 \times 10^{-6} \times 3 \times 10^{8}=4.8 \times 10^{2} V / m $$

Since, we know that for an electromagnetic wave, $\mathbf{E}$ and $\mathbf{B}$ are mutually perpendicular to each other.

So,

$\mathbf{E} \cdot \mathbf{B}=0$

From the given options, when $\hat{\mathbf{n}}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}$

$$ \mathbf{E} \cdot \mathbf{B}=(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}) \cdot(\hat{\mathbf{i}}-2 \hat{\mathbf{j}})=0 $$

Also, when $\hat{\mathbf{n}}=-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}$

$$ \mathbf{E} \cdot \mathbf{B}=(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}) \cdot(-\hat{\mathbf{i}}+2 \hat{\mathbf{j}})=0 $$

But, we also know that the direction of propagation of electromagnetic wave is perpendicular to both $\mathbf{E}$ and $\mathbf{B}$, i.e. it is in the direction of $\mathbf{E} \times \mathbf{B}$.

Again, when $\hat{\mathbf{n}}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}$

$$ \mathbf{E} \times \mathbf{B}=(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}) \times(\hat{\mathbf{i}}-2 \hat{\mathbf{j}})=-\hat{\mathbf{k}} $$

and when $\hat{\mathbf{n}}=-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}$

$$ \mathbf{E} \times \mathbf{B}=(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}) \times(-\hat{\mathbf{i}}+2 \hat{\mathbf{j}})=\hat{\mathbf{k}} $$

But, it is been given in the question that the direction of propagation of wave is in $-\hat{\mathbf{k}}$.

Thus, associated electric field will be

$\mathbf{E}=4.8 \times 10^{2} \cos \left(2 \times 10^{7} z+6 \times 10^{15} t\right)(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}) Vm^{-1}$



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