The magnetic field in the plane electromagnetic field is given by :By=2×10−7sin0.5×103z+1.5×1011tT The expression for the electric field may be given by:

By=2×10−7sin0.5×103z+1.5×1011tTThe electric vector is perpendicular to B as will direction of Propagation of electromagnetic wave. Therefore Ex has to be taken further Eo= B0×c or = 2×10−7×3×108 V/m or =E0=2×10−7×3×108=60 V/m∴ the corresponding value of the electric field is, Ex=60sin0.5×103z+1.5×1011tV/m Hence the correct answer is option (d).

The magnetic field in the plane electromagnetic field is given by :By=2×10−7sin0.5×103z+1.5×1011tT The expression for the electric field may be given by:

By=2×10−7sin0.5×103z+1.5×1011tTThe electric vector is perpendicular to B as will direction of Propagation of electromagnetic wave. Therefore Ex has to be taken further Eo= B0×c or = 2×10−7×3×108 V/m or =E0=2×10−7×3×108=60 V/m∴ the corresponding value of the electric field is, Ex=60sin0.5×103z+1.5×1011tV/m Hence the correct answer is option (d).

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

The magnetic field in the plane electromagnetic field is given by :
$B_{y} = 2 \times 10^{- 7} \sin (0.5 \times 10^{3} z + 1.5 \times 10^{11} t) T$
The expression for the electric field may be given by:

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$E_{x} = 2 \times 10^{- 7} \sin (0.5 \times 10^{3} z + 1.5 \times 10^{11} t) V / m$

$E_{y} = 2 \times 10^{- 7} \sin (0.5 \times 10^{3} z + 1.5 \times 10^{11} t) V / m$

$E_{x} = 60 \sin (0.5 \times 10^{3} z + 1.5 \times 10^{11} t) V / m$

$E_{y} = 60 \sin (0.5 \times 10^{3} z + 1.5 \times 10^{11} t) V / m$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$B_{y} = 2 \times 10^{- 7} \sin (0.5 \times 10^{3} z + 1.5 \times 10^{11} t) T$
The electric vector is perpendicular to B as will direction of Propagation of electromagnetic wave. Therefore E_x has to be taken further E_o= $B_{0} \times c$ or
= $2 \times 10^{- 7} \times 3 \times 10^{8}$ V/m
or = $E_{0} = 2 \times 10^{- 7} \times 3 \times 10^{8} = 60$ V/m
$∴$ the corresponding value of the electric field is, $E_{x} = 60 \sin (0.5 \times 10^{3} z + 1.5 \times 10^{11} t)$ V/m
Hence the correct answer is option (d).