Question

A magnetic field vector in an electromagnetic wave is represented by \[ \vec{B} = B_0 \sin\left(2\pi \nu t - \frac{2\pi x}{\lambda}\right)\hat{j} \] Its associated electric field vector is ________.

• A. \[ \vec{E} = -\nu\lambda B_0 \sin\left(2\pi \nu t - \frac{2\pi x}{\lambda}\right)\hat{k} \]
• B. \[ \vec{E} = -\nu\lambda B_0 \sin\left(2\pi \nu t - \frac{2\pi x}{\lambda}\right)\hat{i} \]
• C. \[ \vec{E} = \nu\lambda B_0 \sin\left(2\pi \nu t - \frac{2\pi x}{\lambda}\right)\hat{k} \]
• D. \[ \vec{E} = \nu\lambda B_0 \sin\left(2\pi \nu t - \frac{2\pi x}{\lambda}\right)\hat{i} \]

Hint:

Remember for electromagnetic waves:

• \[ E_0 = cB_0 \]
• \[ c = \nu\lambda \]
• Direction rule: \[ \vec{E} \times \vec{B} = \text{Propagation direction} \]

Answer 1

The Correct Option is A

Concept: In an electromagnetic wave:

• Electric field \(\vec{E}\), magnetic field \(\vec{B}\), and direction of propagation are mutually perpendicular.
• They satisfy: \[ \vec{E} \times \vec{B} = \text{Direction of propagation} \]
• Magnitudes are related by: \[ E_0 = cB_0 \]

Since: \[ c = \nu\lambda \] we get: \[ E_0 = \nu\lambda B_0 \]

Step 1: Identify the direction of propagation. Given: \[ \vec{B} = B_0\sin\left(2\pi\nu t - \frac{2\pi x}{\lambda}\right)\hat{j} \] The phase: \[ \left(2\pi\nu t - \frac{2\pi x}{\lambda}\right) \] indicates propagation along: \[ +\hat{i} \]

Step 2: Determine direction of electric field. We require: \[ \vec{E} \times \vec{B} = \hat{i} \] Given: \[ \vec{B} \parallel \hat{j} \] Using vector product: \[ (-\hat{k}) \times \hat{j} = \hat{i} \] Hence: \[ \vec{E} \parallel -\hat{k} \]

Step 3: Find magnitude of electric field. Using: \[ E_0 = cB_0 \] and: \[ c = \nu\lambda \] \[ E_0 = \nu\lambda B_0 \] Therefore: \[ \vec{E} = -\nu\lambda B_0 \sin\left(2\pi\nu t - \frac{2\pi x}{\lambda}\right)\hat{k} \] Hence, \[ \boxed{ \vec{E} = -\nu\lambda B_0 \sin\left(2\pi\nu t - \frac{2\pi x}{\lambda}\right)\hat{k} } \]