Calculate the magnetic field due to an electric field E = (75i – 45j) x 10^8 V/m.*1 point(150i - 135j) T(65i - 134j) T(25i – 15j) T(15i – 9j) T

1. Break Down the Problem

We need to calculate the magnetic field $\mathbf{B}$ produced by an electric field $\mathbf{E}$ given by $\mathbf{E} = (75\mathbf{i} - 45\mathbf{j}) \times 10^8 \, \text{V/m}$. We will use the relationship between electric and magnetic fields in certain scenarios, such as in electromagnetic waves where $\mathbf{B}$ can be derived from $\mathbf{E}$.

2. Relevant Concepts

In free space, the magnetic field $\mathbf{B}$ related to the electric field $\mathbf{E}$ with the speed of light $c$ is given by: $\mathbf{B} = \frac{1}{c} \mathbf{E} \times \hat{k}$ where $\hat{k}$ is the unit vector in the direction of propagation (for a plane wave propagating in the z-direction, $\hat{k} = \mathbf{k} = \mathbf{k} \hat{k}$).

3. Analysis and Detail

The magnitude of the speed of light $c$ is approximately $3 \times 10^8 \, \text{m/s}$. The direction of the electric field is in the xy-plane, so we will assume propagation in the z-direction for simplicity:

1. Calculate $\mathbf{E}$: $\mathbf{E} = (75\mathbf{i} - 45\mathbf{j}) \times 10^8 \, \text{V/m}$

2. Calculate $\mathbf{B}$: $\mathbf{B} = \frac{1}{3 \times 10^8} \times (75\mathbf{i} - 45\mathbf{j}) \times 10^8$ $\mathbf{B} = \frac{1}{3} (75\mathbf{i} - 45\mathbf{j})$ $\mathbf{B} = (25\mathbf{i} - 15\mathbf{j}) \, \text{T}$

4. Verify and Summarize

The equation we derived is consistent with the properties of electromagnetic waves and their relation in free space. We substitute the values and confirm that our calculation follows the relationships.

Final Answer

The magnetic field $\mathbf{B}$ is $(25\mathbf{i} - 15\mathbf{j}) \, \text{T}$.