-->

Current Electricity

Question

The oscillating electric field of an electromagnetic wave is given by:
E_y = 30 sin [2 × 10¹¹ t + 300 $π$ x] Vm^-1
Write down the expression for the oscillating magnetic field.

Solution

Oscillating electric field, $E_{y} = 30 \sin [2 \times 10^{11} t + 300 πx] {Vm}^{- 1}$

Here, E_O = 30
Hence, B_o = E_o/c = $\frac{30}{3 \times 10^{8}} = 10^{- 7}$

$∴$ Oscillating magnetic field is given by,

$B_{z} = (10^{- 7}) \sin [2 \times 10^{11} t + 300 πx] T$

For Daily Free Study Material Join wiredfaculty

Current Electricity

The oscillating electric field of an electromagnetic wave is given by:
E_y = 30 sin [2 × 10¹¹ t + 300 $π$ x] Vm^-1
Write down the expression for the oscillating magnetic field.

Some More Questions From Current Electricity Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Loaction

For Daily Free Study Material Join wiredfaculty

Current Electricity

The oscillating electric field of an electromagnetic wave is given by:Ey = 30 sin [2 × 1011 t + 300 πx] Vm-1Write down the expression for the oscillating magnetic field.

Some More Questions From Current Electricity Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

The oscillating electric field of an electromagnetic wave is given by:
E_y = 30 sin [2 × 10¹¹ t + 300 $π$ x] Vm^-1
Write down the expression for the oscillating magnetic field.