SATHEE: Modern Physics 7 Question 82

Modern Physics 7 Question 82

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

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

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

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

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

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Solution:

Given,

$B = 1.6 \times 10^{- 6} \cos (2 \times 10^{7} z + 6 \times 10^{15} t) (2 \hat{i} + \hat{j}) W b m^{- 2}$

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

Equation of associated electric field will be

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

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

So, $| E | = | B | \cdot c$

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

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

So,

$E \cdot B = 0$

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

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

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

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

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

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

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

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

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

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

Thus, associated electric field will be

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

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Modern Physics 7 Question 82

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

Solution:

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चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

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सदस्यता

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Modern Physics 7 Question 82

12. The magnetic field of an electromagnetic wave is given by B=1.6×10−6cos⁡(2×107z+6×1015t)(2i^+j^)Wbm−2

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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