Chapter 22, Problem 46GP

To determine

The value of emf (rms).

Answer to Problem 46GP

The answer is $4.76\times {10}^{-6}\text{V}$ .

Explanation of Solution

Given data:

The rate of the energy flow, $S=1.0\times {10}^{-4}W/m^2$

Speed of the light, $c=3\times {10}^8\text{m/s}$

Permeability of free space, ${\text{μ}}_{\text{0}}=4\pi \times {10}^{-7}\text{T•m/A}$

Frequency, $\text{f}=810\text{kHZ}$

Radius, $r=\frac{2.2}{2}=1.1cm$

Number of loop in coil, $N=380$

Formula used:

It is known that: $S=\frac{1}{2}\left(\frac{c}{{\mu }_0}\right)B_0^2$ and ${\xi }_{\text{0}}=NA{B}_0\omega$

Where,

  $S$ is the power flux.

  $c$ is speed of the light.

  ${\text{μ}}_{\text{0}}$ is permeability of free space

  ${\text{B}}_{\text{0}}$ is the magnetic field.

  ${\xi }_{\text{0}}$ is the maximum electromotive force.

  $\omega$ is angular velocity.

  $A$ is the area of the circular coil.

  $N$ is the number of loop in coil.

Calculation:

Substitute the given data into the formula $S=\frac{1}{2}\left(\frac{c}{{\mu }_0}\right)B_0^2$ to find the magnetic field.

  $B_0=\sqrt{\frac{2\left(1.0\times {10}^{-4}W/m^2\right)\left(4\pi \times {10}^{-7}\text{T•m/A}\right)}{3\times {10}^8\text{m/s}}}=9.15\times {10}^{-10}\text{T}$

Since the given field is oscillates through the coil at $\text{ω}=\text{2πf}$ , so the maximum electromotive force:

  ${\xi }_{\text{0}}=NA{B}_0\omega =\left(380\right)\times \pi \times {\left(0.0011\text{m}\right)}^2\left(9.15\times {10}^{-10}\text{T}\right)\times 2\pi \times \left(810\times {10}^3\text{Hz}\right)=6.73\times {10}^{-6}\text{V}$

The ${\xi }_{rms}=\frac{6.73\times {10}^{-6}\text{V}}{2}=4.76\times {10}^{-6}\text{V}$

Conclusion: The value of root mean square value of magnetic field is $4.76\times {10}^{-6}\text{V}$ .