A circular wire loop has a radius of 7.50 cm. A sinusoidal electromagnetic plane wave traveling in air passes through the loop, with the direction of the magnetic field of the wave perpendicular to the plane of the loop. The intensity of the wave at the location of the loop is 0.0275 W/m2 and the wavelength of the wave is 6.90 m. What is the maximum emf induced in the loop?

The change in magnetic flux through a wire loop induces an emf in the loop and it is given by$\left[\epsilon \right]\mathbf{=}\left|\frac{{\mathrm{d\varphi }}_B}{\mathrm{dt}}\right|$where,${\mathbf{\varphi }}_{\mathbf{B}}\mathbf{=}{\mathbf{B\pi r}}^{\mathbf{2}}$ andThe intensity of a sinusoidal electromagnetic wave in vacuum is related to the electric-field amplitude${\mathbf{E}}_{\mathbf{max}}$magnetic field ${\mathbf{B}}_{\mathbf{max}}$and it is given by equation (32.29) in the form$\mathbf{l}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{\epsilon }}_{\mathbf{0}}{\mathbf{cE}}_{\mathbf{max}}^{\mathbf{2}}$

The sinusoidal electromagnetic wave for the magnetic field is given by$\mathrm{B}=\mathrm{Bcos}{(\mathrm{kx}-\mathrm{\varpi t}+\mathrm{\varphi })}_{\max }$ So,${\mathrm{\varphi }}_{\mathrm{B}}={\mathrm{B\pi r}}^2={\mathrm{\pi r}}^2\mathrm{Bcos}{(\mathrm{kx}-\mathrm{\varpi t})}_{\max }$ Now, plug the expression into equation to get$\epsilon$

$$\begin{gathered} \left[\epsilon \right]=\left|\frac{d{B}_{max}{\mathrm{\pi r}}^2\cos \left(\mathrm{kx}-\mathrm{\varpi t}\right)}{dt}\right| \\ =\varpi B_{max}{\mathrm{\pi r}}^2\sin \left(\mathrm{kx}-\mathrm{\varpi t}\right) \\ ={\mathrm{\varpi B}}_{\max }{\mathrm{\pi r}}^2 \\ =2{\mathrm{\pi fB}}_{\max }{\mathrm{\pi r}}^2 \\ =2{\mathrm{\pi }}^2{\mathrm{fr}}^2{\mathrm{B}}_{\max } \end{gathered}$$

The electromagnetic wave travels with the speed of light and the frequency f of the wave is related to the wavelength and the speed as next $\mathrm{c}=\mathrm{f\lambda }$.Thus can be written as

$$\begin{gathered} \mathrm{f}=\frac{\mathrm{c}}{\mathrm{\lambda }} \\ =\frac{3\times {10}^8\mathrm{m}/\mathrm{s}}{6.9\mathrm{m}}=4.35\times {10}^7\mathrm{Hz} \end{gathered}$$

The intensity of a sinusoidal electromagnetic wave in vacuum is related to the electric-field amplitude E amplitude of magnetic field B and it is given By

$\mathrm{l}=\frac{1}{2}{\mathrm{\epsilon }}_0{\mathrm{E}}_{\max }^2=\frac{1}{2}{\mathrm{\epsilon }}_2\mathrm{c}{\left({\mathrm{cB}}_{\max }\right)}^2=\frac{1}{2}{\mathrm{\epsilon }}_0{\mathrm{c}}^3{\mathrm{B}}_{\max }^2$

Solving for${\mathrm{B}}_{\max }$ we have,

$$\begin{gathered} {\mathrm{B}}_{\max }=\sqrt{\frac{2\mathrm{l}}{{\mathrm{\epsilon }}_0{\mathrm{c}}^3}} \\ =\sqrt{\frac{2\left[\left(0.0275\right)\right]}{\left(8.85\times {10}^{-12}\right)\left(3\times {10}^8\right)}} \\ =1.52\times {10}^{-8}\mathrm{T} \end{gathered}$$

To get$\epsilon$ use equation $\mathrm{\epsilon }=2{\mathrm{\pi }}^2{\mathrm{fr}}^2{\mathrm{B}}_{\max }$.Substitute the values we have,

$$\begin{gathered} \epsilon =2{\mathrm{\pi }}^2\left(4.35\times {10}^7\right)\left(0.007\right)\left(1.52\times {10}^{-8}\right) \\ =73.3\mathrm{mV} \end{gathered}$$

Therefore, the maximum emf induced in the loop is 73.3mV