Question

The magnetic field of a plane electromagnetic wave is given by:

\(\vec B = {B_0}\hat i\left[ {{\rm{cos}}\left( {kz - \omega t} \right)} \right] + {B_1}{\rm{\;}}\hat j{\rm{cos}}\left( {kz + \omega t} \right)\) 

Where B₀ = 3 × 10^-5 T and B₁ = 2 × 10^-6 T

The rms value of the force experienced by a stationary charge Q = 10^-4 C at z = 0 is closest to:
1. 0.6 N
2. 0.1 N
3. 0.9 N
4. 3 × 10^-2 N

Answer 1

Correct Answer - Option 1 : 0.6 N

Concept:

An electromagnetic wave consists of an electric field, defined as usual in terms of the force per charge on a stationary charge, and a magnetic field, defined in terms of the force per charge on a moving charge.

\(\vec B = {B_0}\hat i\left[ {{\rm{cos}}\left( {kz - \omega t} \right)} \right] + {B_1}{\rm{\;}}\hat j{\rm{cos}}\left( {kz + \omega t} \right)\) 

Calculation:

Given,

The magnetic field is given by,

\(\vec B = {B_0}\hat i\left[ {{\rm{cos}}\left( {kz - \omega t} \right)} \right] + {B_1}{\rm{\;}}\hat j{\rm{cos}}\left( {kz + \omega t} \right)\) 

Where, B₀ = 3 × 10^-5 T

B₁ = 2 × 10^-6 T

and Q = 10^-4 C

\(B = \sqrt {B_0^2 + B_1^2} \) 

\(B = \sqrt {{{\left( {30 \times {{10}^{ - 6}}} \right)}^2} + {{\left( {2 \times {{10}^{ - 6}}} \right)}^2}} \) 

\(B = \sqrt {\left( {{{30}^2} + {2^2}} \right)} \times {10^{ - 6}}\) 

B ≈ 30 × 10^-6 T

The energy carried by a wave is proportional to the square of its amplitude (A₂). For E&M waves, it will be (E₀₂), or (B₀₂), or (E₀B₀) where E₀ and B₀ are the maximum values of the electric and magnetic fields intensities.  

Thus, E₀ = cB

∴ E₀ = cB = 3 × 10⁸ × 30 × 10^-6

= 9 × 10³ V/m

Thus, the Root mean square voltage is given by,

\({{\rm{E}}_{{\rm{rms}}}} = \frac{{{E_0}}}{{\sqrt 2 }} = \frac{1}{{\sqrt 2 }} \times 9 \times {10^3}\;V/m\) 

Force on the charge,

\(F = {E_{rms}}Q = \frac{9}{{\sqrt 2 }} \times {10^3} \times {10^{ - 4}} \simeq 0.64N\) 

Thus, the rms value of the force experienced by the charge is 0.6 N.