Q42. A plane electromagnetic wave of frequency 30MHz travels in free space along the x-direction. At

Q: What is the correct answer for CUET UG Physics Q42?

The correct answer is a. For a plane electromagnetic wave traveling in free space:1. The speed of the wave (c) is related to the magnitudes of the electric field (E) and magnetic field (B) by c = E/B. Thus, B = E/c.2. The electric field (E), magnetic field (B), and the direction of propagation (**k**) are mutually perpendicular and form a right-handed system (i.e., **k** = **E** × **B**).Given:- Speed of light in free

Q: Q42. A plane electromagnetic wave of frequency 30MHz travels in free space along the x-direction. At a particular point in space and time E = 6.3ĵ V/m. The B at this point would be: (a) 2.1 × 10⁻⁸ k̂…

The correct answer is a. For a plane electromagnetic wave traveling in free space:1. The speed of the wave (c) is related to the magnitudes of the electric field (E) and magnetic field (B) by c = E/B. Thus, B = E/c.2. The electric field (E), magnetic field (B), and the direction of propagation (**k**) are mutually perpendicular and form a right-handed system (i.e., **k** = **E** × **B**).Given:- Speed of light in free

Q: What are the options for this CUET UG Physics question?

A. 2.1 × 10⁻⁸ k̂ T · B. 2.1 x 10⁻⁶ T · C. 1.6 × 10⁻⁸ î T · D. 2.6 x 10⁻⁸ T

Question

Q42. A plane electromagnetic wave of frequency 30MHz travels in free space along the x-direction. At a particular point in space and time E = 6.3ĵ V/m. The B at this point would be:
(a) 2.1 × 10⁻⁸ k̂ T
(b) 2.1 x 10⁻⁶ T
(c) 1.6 × 10⁻⁸ î T
(d) 2.6 x 10⁻⁸ T

FirstInTest · Accepted Answer

For a plane electromagnetic wave traveling in free space:1.  The speed of the wave (c) is related to the magnitudes of the electric field (E) and magnetic field (B) by c = E/B. Thus, B = E/c.2.  The electric field (E), magnetic field (B), and the direction of propagation (**k**) are mutually perpendicular and form a right-handed system (i.e., **k** = **E** × **B**).Given:-   Speed of light in free space (c) = 3 × 10⁸ m/s-   Electric field at a point **E** = 6.3 ĵ V/m (so, E = 6.3 V/m, and direction is +y axis)-   Direction of propagation is along the x-direction (î).First, calculate the magnitude of the magnetic field (B):B = E / c = (6.3 V/m) / (3 × 10⁸ m/s) = 2.1 × 10⁻⁸ T.Next, determine the direction of the magnetic field (B):We have **k** (propagation) along î, and **E** along ĵ. We need to find the direction of **B** such that î = ĵ × **B**.According to the right-hand rule, ĵ × k̂ = î. Therefore, **B** must be along the +z axis (k̂).So, the magnetic field at that point would be **B** = 2.1 × 10⁻⁸ k̂ T.

Answer

2.1 × 10⁻⁸ k̂ T

Answer

2.1 x 10⁻⁶ T

Answer

1.6 × 10⁻⁸ î T

Answer

2.6 x 10⁻⁸ T

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