The given physical quantity is defined as:
\( \vec{S} = \frac{\vec{E} \times \vec{B}}{\mu_0} \)
Where \( \vec{E} \) is the electric field, \( \vec{B} \) is the magnetic field, and \( \mu_0 \) is the permeability of free space. We are asked to find the dimensions of \( \vec{S} \) and compare it to the dimensions of other physical quantities.
Step 1: Analyzing the dimensions of each term
The electric field \( \vec{E} \) has dimensions of force per unit charge, which can be expressed as:
\( [E] = \frac{ML}{T^3I} \)
The magnetic field \( \vec{B} \) has dimensions of force per unit charge per velocity, which can be written as:
\( [B] = \frac{M}{T^2I} \)
The permeability of free space \( \mu_0 \) has dimensions:
\( [\mu_0] = \frac{M}{A^2T^2} \)
Step 2: Calculating the dimensions of \( \vec{S} \)
The dimensions of \( \vec{S} \) are given by:
\( [\vec{S}] = \frac{[E] \times [B]}{[\mu_0]} \)
Substituting the dimensions of \( \vec{E} \), \( \vec{B} \), and \( \mu_0 \):
\( [\vec{S}] = \frac{\left( \frac{ML}{T^3I} \right) \times \left( \frac{M}{T^2I} \right)}{\frac{M}{A^2T^2}} = \frac{M^2L}{T^5I^2} \times \frac{A^2T^2}{M} = \frac{ML^2}{T^3I^2A^2} \)
Step 3: Identifying the quantity with the same dimensions
The dimensions of \( \vec{S} \) are the same as the dimensions of:
\( \frac{\text{Power}}{\text{Area}} \)
This is because power has dimensions of \( \frac{ML^2}{T^3I} \) and area has dimensions of \( L^2 \), so \( \frac{\text{Power}}{\text{Area}} \) has the same dimensions as \( \vec{S} \).
Final Answer:
\( \frac{\text{Power}}{\text{Area}} \)
The dimension of $ \sqrt{\frac{\mu_0}{\epsilon_0}} $ is equal to that of: (Where $ \mu_0 $ is the vacuum permeability and $ \epsilon_0 $ is the vacuum permittivity)
The unit of $ \sqrt{\frac{2I}{\epsilon_0 c}} $ is: (Where $ I $ is the intensity of an electromagnetic wave, and $ c $ is the speed of light)
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): Electromagnetic waves carry energy but not momentum.
Reason (R): Mass of a photon is zero. In the light of the above statements.
choose the most appropriate answer from the options given below:
Let $ y(x) $ be the solution of the differential equation $$ x^2 \frac{dy}{dx} + xy = x^2 + y^2, \quad x > \frac{1}{e}, $$ satisfying $ y(1) = 0 $. Then the value of $ 2 \cdot \frac{(y(e))^2}{y(e^2)} $ is ________.
The waves that are produced when an electric field comes into contact with a magnetic field are known as Electromagnetic Waves or EM waves. The constitution of an oscillating magnetic field and electric fields gives rise to electromagnetic waves.
Electromagnetic waves can be grouped according to the direction of disturbance in them and according to the range of their frequency. Recall that a wave transfers energy from one point to another point in space. That means there are two things going on: the disturbance that defines a wave, and the propagation of wave. In this context the waves are grouped into the following two categories: