PhysicsThe magnetic field of a plane electromagnetic wave is given by B→=2×10−8sin⁡(0.5×103x+1.5×1011t)j^T. The amplitude of the electric field would be

The magnetic field of a plane electromagnetic wave is given by $\vec{B} = 2 \times 10^{- 8} \sin (0 .5 \times 10^{3} x + 1 .5 \times 10^{11} t) \hat{j} T$ . The amplitude of the electric field would be

A
$6 V / m along x -axis$
B
$3 V / m along z -axis$
C
$6 V / m along z -axis$
D
$2 \times 10^{- 8} V / m along z -axis$

Solution:

$| \vec{E} | = c | \vec{B} | = 3 \times 10^{8} \times 2 \times 10^{- 8}$

$| \vec{E} | = 6 V / m along z-axis$

Related content

Sine and Cosine Waves

Zener Diode

Hygrometer

Half Wave Rectifier

Compound Microscope

Solar Cooker

MCB

Difference Between Scalar and Vector Quantity

Difference Between Interference and Diffraction

Difference Between Heat and Temperature

The magnetic field of a plane electromagnetic wave is given by $\vec{B} = 2 \times 10^{- 8} \sin (0 .5 \times 10^{3} x + 1 .5 \times 10^{11} t) \hat{j} T$ . The amplitude of the electric field would be

Solution:

Related content

Company

Support

Courses

More

JEE

NEET

CUET

CBSE

Popular books

NCERT solutions

NCERT exemplar

State Board

Subject

follow us