To find the magnetic field corresponding to the given electric field of an electromagnetic wave, we use the relationship between the electric and magnetic fields in a plane electromagnetic wave. The relationship is given by:
\(E = cB\),
where:
Given:
Since the wave is traveling in the X-direction, the magnetic field will be perpendicular to both the direction of wave propagation (X-direction) and the electric field (Y-direction). Therefore, it must be in the Z-direction. Let's calculate the magnitude of the magnetic field:
\(B = \frac{E}{c} = \frac{9.6}{3 \times 10^8}\)
Calculating the value:
\(B = 3.2 \times 10^{-8} \, \text{T}\)
Since the magnetic field is in the Z-direction, we write the magnetic field as:
\(\vec{B} = 3.2 \times 10^{-8} \hat{k} \, \text{T}\)
Thus, the correct option is:
\(3.2 \times 10^{-8} \hat{k} \, \text{T}\)
Let's rule out other options:
Therefore, the value of the magnetic field at this point is \(3.2 \times 10^{-8} \hat{k} \, \text{T}\).
For an electromagnetic wave, the electric field \( \vec{E} \) and magnetic field \( \vec{B} \) are related by the speed of light \( c \):
\(\frac{E}{B} = c\)
where \( c = 3 \times 10^8 \, \text{m/s} \).
Step 1: Calculate the magnetic field \( B \):
Given \( E = 9.6 \, \text{V/m} \),
\(B = \frac{E}{c} = \frac{9.6}{3 \times 10^8} = 3.2 \times 10^{-8} \, \text{T}\)
Step 2: Determine the direction of \( \vec{B} \):
Since the wave travels along the X-direction and \( \vec{E} \) is along \( \hat{j} \), the magnetic field \( \vec{B} \) must be perpendicular to both the direction of propagation and \( \vec{E} \). By the right-hand rule:
\(\vec{B} \, \text{points along} \, \hat{k}.\)
Thus, the magnetic field at this point is \( 3.2 \times 10^{-8} \hat{k} \, \text{T}. \)
The Correct Answer is: \( 3.2 \times 10^{-8} \hat{k} \, \text{T}. \)
A 2 $\text{kg}$ mass is attached to a spring with spring constant $ k = 200, \text{N/m} $. If the mass is displaced by $ 0.1, \text{m} $, what is the potential energy stored in the spring?
| List-I | List-II |
|---|---|
| (A) A force that restores an elastic body of unit area to its original state | (I) Bulk modulus |
| (B) Two equal and opposite forces parallel to opposite faces | (IV) Shear modulus |
| (C) Forces perpendicular everywhere to the surface per unit area same everywhere | (III) Stress |
| (D) Two equal and opposite forces perpendicular to opposite faces | (II) Young's modulus |

