PYQs Homework Notes Mocks Board Daily Homework JEE Mains Physics Question type Equation of B from the equation of E (or the reverse) for an EM wave This type of question, asked 10 times: 26 Jun 2022, 28 Jun 2022, 30 Jan 2024, 4 Apr 2024, 9 Apr 2024, 23 Jan 2025, 28 Jan 2025, 21 Jan 2026 S1, 28 Jan 2026 S2, 4 Apr 2026 S2. All 11 questions below, oldest first.
Try each one first, then open its answer.
Question 1 26 Jun 2022
15 If electric field intensity of a uniform plane electromagnetic wave is given as E = − 301.6 sin ( k z − ω t ) a ^ x + 452.4 sin ( k z − ω t ) E = - 301.6\sin(kz - \omega t){\widehat{a}}_{x}+ 452.4\sin(kz - \omega t) E = − 301.6 sin ( k z − ω t ) a x + 452.4 sin ( k z − ω t ) a ^ y V m {\widehat{a}}_{y}\frac{V}{m} a y m V Then, magnetic intensity H H H of this wave in A m − 1 Am^{- 1} A m − 1 will be:' [Given: Speed of light in vacuum c = 3 × 10 8 m s − 1 = 3 \times10^{8}{\text{ }ms}^{- 1} = 3 × 1 0 8 m s − 1 , permeability of vacuum μ 0 = 4 π × 10 − 7 N A − 2 \mu_{0}= 4\pi \times10^{- 7}NA^{- 2} μ 0 = 4 π × 1 0 − 7 N A − 2 ] A. + 0.8 sin ( k z − ω t ) a ^ y + 0.8 sin ( k z − ω t ) a ^ x + 0.8\sin(kz - \omega t){\widehat{a}}_{y} + 0.8\sin(kz - \omega t){\widehat{a}}_{x} + 0.8 sin ( k z − ω t ) a y + 0.8 sin ( k z − ω t ) a x B. + 1.0 × 10 − 6 sin ( k z − ω t ) a ^ y + 1.5 × 10 − 6 ( k z − ω t ) a ^ x + 1.0 \times 10^{- 6}\sin(kz - \omega t){\widehat{a}}_{y} + 1.5 \times 10^{- 6}(kz - \omega t){\widehat{a}}_{x} + 1.0 × 1 0 − 6 sin ( k z − ω t ) a y + 1.5 × 1 0 − 6 ( k z − ω t ) a x C. − 0.8 sin ( k z − ω t ) a ^ y − 1.2 sin ( k z − ω t ) a ^ x - 0.8\sin(kz - \omega t){\widehat{a}}_{y} - 1.2\sin(kz - \omega t){\widehat{a}}_{x} − 0.8 sin ( k z − ω t ) a y − 1.2 sin ( k z − ω t ) a x D. − 1.0 × 10 − 6 sin ( k z − ω t ) a ^ y − 1.5 × 10 − 6 sin ( k z − ω t ) a ^ x - 1.0 \times 10^{- 6}\sin(kz - \omega t){\widehat{a}}_{y} - 1.5 \times 10^{- 6}\sin(kz - \omega t){\widehat{a}}_{x} − 1.0 × 1 0 − 6 sin ( k z − ω t ) a y − 1.5 × 1 0 − 6 sin ( k z − ω t ) a x Tap an option to check your answer.
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Question 2 28 Jun 2022
105 An EM wave propagating in x x x -direction has a wavelength of 8 m m 8\text{ }mm 8 mm . The electric field vibrating in the y-direction has maximum magnitude of 60 V m − 1 60Vm^{- 1} 60 V m − 1 . Choose the correct equations for electric and magnetic fields if the EM wave is propagating in vacuum: A. E y = 60 sin [ π 4 × 10 3 ( x − 3 × 10 8 t ) ] j ^ V m − 1 E_{y} = 60\sin\left\lbrack \frac{\pi}{4} \times 10^{3}\left( x - 3 \times 10^{8}t \right) \right\rbrack\widehat{j}\ Vm^{- 1} E y = 60 sin [ 4 π × 1 0 3 ( x − 3 × 1 0 8 t ) ] j V m − 1 , B z = 2 sin [ π 4 × 10 3 ( x − 3 × 10 8 t ) ] k ^ T B_{z} = 2\sin\left\lbrack \frac{\pi}{4} \times 10^{3}\left( x - 3 \times 10^{8}t \right) \right\rbrack\widehat{k}\ T B z = 2 sin [ 4 π × 1 0 3 ( x − 3 × 1 0 8 t ) ] k T B. E y = 60 sin [ π 4 × 10 3 ( x − 3 × 10 8 t ) ] j ^ V m − 1 E_{y} = 60\sin\left\lbrack \frac{\pi}{4} \times 10^{3}\left( x - 3 \times 10^{8}t \right) \right\rbrack\widehat{j}\ Vm^{- 1} E y = 60 sin [ 4 π × 1 0 3 ( x − 3 × 1 0 8 t ) ] j V m − 1 , B z = 2 × 10 − 7 sin [ π 4 × 10 3 ( x − 3 × 10 8 t ) ] k ^ T B_{z} = 2 \times 10^{- 7}\sin\left\lbrack \frac{\pi}{4} \times 10^{3}\left( x - 3 \times 10^{8}t \right) \right\rbrack\widehat{k}\ T B z = 2 × 1 0 − 7 sin [ 4 π × 1 0 3 ( x − 3 × 1 0 8 t ) ] k T C. E y = 2 × 10 − 7 sin [ π 4 × 10 3 ( x − 3 × 10 8 t ) ] j ^ V m − 1 E_{y} = 2 \times 10^{- 7}\sin\left\lbrack \frac{\pi}{4} \times 10^{3}\left( x - 3 \times 10^{8}t \right) \right\rbrack\widehat{j}\ Vm^{- 1} E y = 2 × 1 0 − 7 sin [ 4 π × 1 0 3 ( x − 3 × 1 0 8 t ) ] j V m − 1 , B z = 60 sin [ π 4 × 10 3 ( x − 3 × 10 8 t ) ] k ^ T B_{z} = 60\sin\left\lbrack \frac{\pi}{4} \times 10^{3}\left( x - 3 \times 10^{8}t \right) \right\rbrack\widehat{k}\ T B z = 60 sin [ 4 π × 1 0 3 ( x − 3 × 1 0 8 t ) ] k T D. E y = 2 × 10 − 7 sin [ π 4 × 10 4 ( x − 4 × 10 8 t ) ] j ^ V m − 1 E_{y} = 2 \times 10^{- 7}\sin\left\lbrack \frac{\pi}{4} \times 10^{4}\left( x - 4 \times 10^{8}t \right) \right\rbrack\widehat{j}\ Vm^{- 1} E y = 2 × 1 0 − 7 sin [ 4 π × 1 0 4 ( x − 4 × 1 0 8 t ) ] j V m − 1 , B z = 60 sin [ π 4 × 10 4 ( x − 4 × 10 8 t ) ] k ^ T B_{z} = 60\sin\left\lbrack \frac{\pi}{4} \times 10^{4}\left( x - 4 \times 10^{8}t \right) \right\rbrack\widehat{k}\ T B z = 60 sin [ 4 π × 1 0 4 ( x − 4 × 1 0 8 t ) ] k T Tap an option to check your answer.
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Question 3 30 Jan 2024
8 The electric field of an electromagnetic wave in free space is represented as E → = E 0 cos ( ω t − k z ) i ^ \overrightarrow{E}=E_{0}\cos(\omega t - kz)\widehat{i} E = E 0 cos ( ω t − k z ) i . The corresponding magnetic induction vector will be: A. B → = E 0 C c o s ( ω t − k z ) j ^ \overrightarrow{B} = E_{0}Ccos(\omega t - kz)\widehat{j} B = E 0 C cos ( ω t − k z ) j B. B → = E 0 C cos ( ω t − k z ) j ^ \overrightarrow{B} = \frac{E_{0}}{C}\cos(\omega t - kz)\widehat{j} B = C E 0 cos ( ω t − k z ) j C. B → = E 0 C o s ( ω t + k z ) j ^ \overrightarrow{B} = E_{0}Cos(\omega t + kz)\widehat{j} B = E 0 C os ( ω t + k z ) j D. B → = E 0 C cos ( ω t + k z ) j ^ \overrightarrow{B} = \frac{E_{0}}{C}\cos(\omega t + kz)\widehat{j} B = C E 0 cos ( ω t + k z ) j Tap an option to check your answer.
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Question 4 4 Apr 2024
2 The electric field in an electromagnetic wave is given by E → = i ^ 40 cos ω ( t − z c ) N C − 1 \overrightarrow{E}=\widehat{i}40\cos\omega\left( t -\frac{z}{c} \right)NC^{- 1} E = i 40 cos ω ( t − c z ) N C − 1 . The magnetic field induction of this wave is (in SI unit): A. B → = i ^ 40 c cos ω ( t − z c ) \overrightarrow{B} = \widehat{i}\frac{40}{c}\cos\omega\left( t - \frac{z}{c} \right) B = i c 40 cos ω ( t − c z ) B. B → = j ^ 40 cos ω ( t − z c ) \overrightarrow{B} = \widehat{j}40\cos\omega\left( t - \frac{z}{c} \right) B = j 40 cos ω ( t − c z ) C. B → = k ^ 40 c cos ω ( t − Z c ) \overrightarrow{B} = \widehat{k}\frac{40}{c}\cos\omega\left( t - \frac{Z}{c} \right) B = k c 40 cos ω ( t − c Z ) D. B → = j ^ 40 c cos ω ( t − z c ) \overrightarrow{B} = \widehat{j}\frac{40}{c}\cos\omega\left( t - \frac{z}{c} \right) B = j c 40 cos ω ( t − c z ) Tap an option to check your answer.
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Question 5 9 Apr 2024
3 A plane EM wave is propagating along x x x direction. It has a wavelength of 4 m m 4\text{ }mm 4 mm . If electric field is in y y y direction with the maximum magnitude of 60 V m − 1 60Vm^{- 1} 60 V m − 1 , the equation for magnetic field is: A. B z = 60 sin [ π 2 ( x − 3 × 10 8 t ) ] k ^ T B_{z} = 60\sin\left\lbrack \frac{\pi}{2}\left( x - 3 \times 10^{8}t \right) \right\rbrack\widehat{k}\ T B z = 60 sin [ 2 π ( x − 3 × 1 0 8 t ) ] k T B. B z = 2 × 10 − 7 sin [ π 2 × 10 3 ( x − 3 × 10 8 t ) ] k ^ T B_{z} = 2 \times 10^{- 7}\sin\left\lbrack \frac{\pi}{2} \times 10^{3}\left( x - 3 \times 10^{8}t \right) \right\rbrack\widehat{k}\ T B z = 2 × 1 0 − 7 sin [ 2 π × 1 0 3 ( x − 3 × 1 0 8 t ) ] k T C. B x = 60 sin [ π 2 ( x − 3 × 10 8 t ) ] i ^ T B_{x} = 60\sin\left\lbrack \frac{\pi}{2}\left( x - 3 \times 10^{8}t \right) \right\rbrack\widehat{i}T B x = 60 sin [ 2 π ( x − 3 × 1 0 8 t ) ] i T D. B z = 2 × 10 − 7 sin [ π 2 ( x − 3 × 10 8 t ) ] k ^ T B_{z} = 2 \times 10^{- 7}\sin\left\lbrack \frac{\pi}{2}\left( x - 3 \times 10^{8}t \right) \right\rbrack\widehat{k}T B z = 2 × 1 0 − 7 sin [ 2 π ( x − 3 × 1 0 8 t ) ] k T Tap an option to check your answer.
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Question 6 9 Apr 2024
95 The magnetic field in a plane electromagnetic wave is B y = ( 3.5 × 10 − 7 ) sin ( 1.5 × 10 3 x + 0.5 × 10 11 t ) T B_{y}=\left( 3.5 \times 10^{- 7} \right)\sin\left( 1.5 \times 10^{3}x + 0.5 \times 10^{11}t \right)T B y = ( 3.5 × 1 0 − 7 ) sin ( 1.5 × 1 0 3 x + 0.5 × 1 0 11 t ) T . The corresponding electric field will be A. E y = 1.17 sin ( 1.5 × 10 3 x + 0.5 × 10 11 t ) V m − 1 E_{y} = 1.17\sin\left( 1.5 \times 10^{3}x + 0.5 \times 10^{11}t \right){Vm}^{- 1} E y = 1.17 sin ( 1.5 × 1 0 3 x + 0.5 × 1 0 11 t ) V m − 1 B. E z = 105 sin ( 1.5 × 10 3 x + 0.5 × 10 11 t ) V m − 1 E_{z} = 105\sin\left( 1.5 \times 10^{3}x + 0.5 \times 10^{11}t \right){Vm}^{- 1} E z = 105 sin ( 1.5 × 1 0 3 x + 0.5 × 1 0 11 t ) V m − 1 C. E z = 1.17 sin ( 1.5 × 10 3 x + 0.5 × 10 11 t ) V m − 1 E_{z} = 1.17\sin\left( 1.5 \times 10^{3}x + 0.5 \times 10^{11}t \right){Vm}^{- 1} E z = 1.17 sin ( 1.5 × 1 0 3 x + 0.5 × 1 0 11 t ) V m − 1 D. E y = 10.5 sin ( 1.5 × 10 3 x + 0.5 × 10 11 t ) V m − 1 E_{y} = 10.5\sin\left( 1.5 \times 10^{3}x + 0.5 \times 10^{11}t \right){Vm}^{- 1} E y = 10.5 sin ( 1.5 × 1 0 3 x + 0.5 × 1 0 11 t ) V m − 1 Tap an option to check your answer.
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Question 7 23 Jan 2025
18 The electric field of an electromagnetic wave in free space is E → = 57 cos [ 7.5 × 10 6 t − 5 × 10 − 3 ( 3 x + 4 y ) ] ( 4 i ^ − 3 j ^ ) N / C \begin{matrix} & \overrightarrow{E}= 57\cos\left\lbrack 7.5 \times10^{6}t - 5 \times10^{- 3}(3x + 4y) \right\rbrack \\ & \ (4\widehat{i}- 3\widehat{j})N/C \end{matrix} E = 57 cos [ 7.5 × 1 0 6 t − 5 × 1 0 − 3 ( 3 x + 4 y ) ] ( 4 i − 3 j ) N / C The associated magnetic field in Tesla is- A. B → = 57 3 × 10 8 cos [ 7.5 × 10 6 t − 5 × 10 − 3 ( 3 x + 4 y ) ] ( 5 k ^ ) \overrightarrow{B} = \frac{57}{3 \times 10^{8}}\cos\left\lbrack 7.5 \times 10^{6}t - 5 \times 10^{- 3}(3x + 4y) \right\rbrack(5\widehat{k}) B = 3 × 1 0 8 57 cos [ 7.5 × 1 0 6 t − 5 × 1 0 − 3 ( 3 x + 4 y ) ] ( 5 k ) B. B → = 57 3 × 10 8 cos [ 7.5 × 10 6 t − 5 × 10 − 3 ( 3 x + 4 y ) ] ( k ^ ) \overrightarrow{B} = \frac{57}{3 \times 10^{8}}\cos\left\lbrack 7.5 \times 10^{6}t - 5 \times 10^{- 3}(3x + 4y) \right\rbrack(\widehat{k}) B = 3 × 1 0 8 57 cos [ 7.5 × 1 0 6 t − 5 × 1 0 − 3 ( 3 x + 4 y ) ] ( k ) C. B → = − 57 3 × 10 8 cos [ 7.5 × 10 6 t − 5 × 10 − 3 ( 3 x + 4 y ) ] ( 5 k ^ ) \overrightarrow{B} = - \frac{57}{3 \times 10^{8}}\cos\left\lbrack 7.5 \times 10^{6}t - 5 \times 10^{- 3}(3x + 4y) \right\rbrack(5\widehat{k}) B = − 3 × 1 0 8 57 cos [ 7.5 × 1 0 6 t − 5 × 1 0 − 3 ( 3 x + 4 y ) ] ( 5 k ) D. B → = − 57 3 × 10 8 cos [ 7.5 × 10 6 t − 5 × 10 − 3 ( 3 x + 4 y ) ] ( k ^ ) \overrightarrow{B} = - \frac{57}{3 \times 10^{8}}\cos\left\lbrack 7.5 \times 10^{6}t - 5 \times 10^{- 3}(3x + 4y) \right\rbrack(\widehat{k}) B = − 3 × 1 0 8 57 cos [ 7.5 × 1 0 6 t − 5 × 1 0 − 3 ( 3 x + 4 y ) ] ( k ) Tap an option to check your answer.
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Question 8 28 Jan 2025
94 The magnetic field of an E.M. wave is given by B → = ( 3 2 i ^ + 1 2 j ^ ) 30 sin [ ω ( t − z c ) ] \overrightarrow{B}=\left( \frac{\sqrt{3}}{2}\widehat{i}+\frac{1}{2}\widehat{j} \right)30\sin\left\lbrack \omega\left( t -\frac{z}{c} \right) \right\rbrack B = ( 2 3 i + 2 1 j ) 30 sin [ ω ( t − c z ) ] (S.I. Units) The corresponding electric field in S.I. units is : A. E → = ( 1 2 i ^ − 3 2 j ^ ) 30 c sin [ ω ( t − z c ) ] \overrightarrow{E} = \left( \frac{1}{2}\widehat{i} - \frac{\sqrt{3}}{2}\widehat{j} \right)30c\ \sin\left\lbrack \omega\left( t - \frac{z}{c} \right) \right\rbrack E = ( 2 1 i − 2 3 j ) 30 c sin [ ω ( t − c z ) ] B. E → = ( 3 4 i ^ + 1 4 j ^ ) 30 c cos [ ω ( t − z c ) ] \overrightarrow{E} = \left( \frac{3}{4}\widehat{i} + \frac{1}{4}\widehat{j} \right)30c\ \cos\left\lbrack \omega\left( t - \frac{z}{c} \right) \right\rbrack E = ( 4 3 i + 4 1 j ) 30 c cos [ ω ( t − c z ) ] C. E → = ( 1 2 i ^ + 3 2 j ^ ) 30 c sin [ ω ( t + z c ) ] \overrightarrow{E} = \left( \frac{1}{2}\widehat{i} + \frac{\sqrt{3}}{2}\widehat{j} \right)30c\ \sin\left\lbrack \omega\left( t + \frac{z}{c} \right) \right\rbrack E = ( 2 1 i + 2 3 j ) 30 c sin [ ω ( t + c z ) ] D. E → = ( 3 2 i ^ − 1 2 j ^ ) 30 c sin [ ω ( t + z c ) ] \overrightarrow{E} = \left( \frac{\sqrt{3}}{2}\widehat{i} - \frac{1}{2}\widehat{j} \right)30c\ \sin\left\lbrack \omega\left( t + \frac{z}{c} \right) \right\rbrack E = ( 2 3 i − 2 1 j ) 30 c sin [ ω ( t + c z ) ] Tap an option to check your answer.
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Question 9 21 Jan 2026 S1
10 The electric field a plane electromagnetic wave is given by : E y = 69 sin [ 0.6 × 10 3 x − 1.8 × 10 11 t ] V / m E_{y}= 69\sin\left\lbrack 0.6 \times10^{3}x - 1.8 \times10^{11}t \right\rbrack V/m E y = 69 sin [ 0.6 × 1 0 3 x − 1.8 × 1 0 11 t ] V / m . The expression for magnetic field associated with this electromagnetic wave is _ _ _ _ \_\_\_\_ ____ T. A. B z = 2.3 × 10 − 7 sin [ 0.6 × 10 3 x − 1.8 × 10 11 t ] B_{z} = 2.3 \times 10^{- 7}\sin\left\lbrack 0.6 \times 10^{3}x - 1.8 \times 10^{11}t \right\rbrack B z = 2.3 × 1 0 − 7 sin [ 0.6 × 1 0 3 x − 1.8 × 1 0 11 t ] B. B z = 2.3 × 10 − 7 sin [ 0.6 × 10 3 x + 1.8 × 10 11 t ] B_{z} = 2.3 \times 10^{- 7}\sin\left\lbrack 0.6 \times 10^{3}x + 1.8 \times 10^{11}t \right\rbrack B z = 2.3 × 1 0 − 7 sin [ 0.6 × 1 0 3 x + 1.8 × 1 0 11 t ] C. B y = 69 sin [ 0.6 × 10 3 x + 1.8 × 10 11 t ] B_{y} = 69\sin\left\lbrack 0.6 \times 10^{3}x + 1.8 \times 10^{11}t \right\rbrack B y = 69 sin [ 0.6 × 1 0 3 x + 1.8 × 1 0 11 t ] D. B y = 2.3 × 10 − 7 sin [ 0.6 × 10 3 x − 1.8 × 10 11 t ] B_{y} = 2.3 \times 10^{- 7}\sin\left\lbrack 0.6 \times 10^{3}x - 1.8 \times 10^{11}t \right\rbrack B y = 2.3 × 1 0 − 7 sin [ 0.6 × 1 0 3 x − 1.8 × 1 0 11 t ] Tap an option to check your answer.
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Question 10 28 Jan 2026 S2
2 A plane electromagnetic wave is moving in free space with velocity c = 3 × 10 8 m / s c = 3 \times10^{8}\text{ }m/s c = 3 × 1 0 8 m / s and its electric field is given as E → = 54 sin ( k z − ω t ) j ^ V / m \overrightarrow{E}= 54\sin(kz -\omega t)\widehat{j}V/m E = 54 sin ( k z − ω t ) j V / m , where j ^ \widehat{j} j is the unit vector along y y y -axis. The magnetic field vector B → \overrightarrow{B} B of the wave is : A. − 1.8 × 10 − 7 sin ( k z − ω t ) i ^ T - 1.8 \times 10^{- 7}\sin(kz - \omega t)\widehat{i}\ T − 1.8 × 1 0 − 7 sin ( k z − ω t ) i T B. 1.4 × 10 − 7 sin ( k z − ω t ) k ^ T 1.4 \times 10^{- 7}\sin(kz - \omega t)\widehat{k}T 1.4 × 1 0 − 7 sin ( k z − ω t ) k T C. 1.4 × 10 − 7 sin ( k z − ω t ) i ^ T 1.4 \times 10^{- 7}\sin(kz - \omega t)\widehat{i}T 1.4 × 1 0 − 7 sin ( k z − ω t ) i T D. + 1.8 × 10 − 7 sin ( k z − ω t ) i ^ T + 1.8 \times 10^{- 7}\sin(kz - \omega t)\widehat{i}\ T + 1.8 × 1 0 − 7 sin ( k z − ω t ) i T Tap an option to check your answer.
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Question 11 4 Apr 2026 S2
15 A magnetic field vector in an electromagnetic wave is represented by, B → = B 0 sin ( 2 π v t − 2 π x λ ) j ^ \overrightarrow{B}=B_{0}\sin\left( 2\pi vt-\frac{2\pi x}{\lambda} \right)\widehat{j} B = B 0 sin ( 2 π v t − λ 2 π x ) j . Its associated electric field vector is _ _ _ _ \_\_\_\_ ____ . A. E → = − v λ B 0 sin ( 2 π v t − 2 π x λ ) k ^ \overrightarrow{E} = - v\lambda B_{0}\sin\left( 2\pi vt - \frac{2\pi x}{\lambda} \right)\widehat{k} E = − v λ B 0 sin ( 2 π v t − λ 2 π x ) k B. E → = − v λ B 0 sin ( 2 π v t − 2 π x λ ) i ^ \overrightarrow{E} = - v\lambda B_{0}\sin\left( 2\pi vt - \frac{2\pi x}{\lambda} \right)\widehat{i} E = − v λ B 0 sin ( 2 π v t − λ 2 π x ) i C. E → = v λ B 0 sin ( 2 π v t − 2 π x λ ) k ^ \overrightarrow{E} = v\lambda{\text{ }B}_{0}\sin\left( 2\pi vt - \frac{2\pi x}{\lambda} \right)\widehat{k} E = v λ B 0 sin ( 2 π v t − λ 2 π x ) k D. E → = v λ B 0 sin ( 2 π v t − 2 π x λ ) i ^ \overrightarrow{E} = v\lambda{\text{ }B}_{0}\sin\left( 2\pi vt - \frac{2\pi x}{\lambda} \right)\widehat{i} E = v λ B 0 sin ( 2 π v t − λ 2 π x ) i Tap an option to check your answer.
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