Step 1
(a) The given value represents the intensity \(\displaystyle{I}={1.4}{k}\frac{{W}}{{m}^{{{2}}}}\) of the light. The intensity of a sinusoidal electromagnetic wave in vacuum is related to the electric-field amplitude \(\displaystyle{E}_{{\max}}\) and the amplitude of magnetic field \(\displaystyle{B}_{{\max}}\) and it is given by equation (32.29) in the form

\(\displaystyle{I}={\frac{{{1}}}{{{2}}}}\epsilon_{{{o}}}{c}{{E}_{{\max}}^{{{2}}}}\) (1)

Where \(\displaystyle\epsilon_{{{o}}}\) is the electric constant, cc is the speed of light. Solve equation (1) for \(\displaystyle{E}_{{\max}}\)

\(\displaystyle{E}_{{\max}}=\sqrt{{{\frac{{{2}{I}}}{{\epsilon_{{{o}}}{c}}}}}}\) (2)

Plug the values for I, c, and \(\displaystyle\epsilon_{{{o}}}\) into equation (2) to get \(\displaystyle{E}_{{\max}}\)

\(\displaystyle{E}_{{\max}}=\sqrt{{{\frac{{{2}{I}}}{{\epsilon_{{{o}}}{c}}}}}}\)

\(\displaystyle=\sqrt{{{\frac{{{2}{\left({1400}\frac{{W}}{{m}^{{{2}}}}\right)}}}{{{\left({8.85}\times{10}^{{-{12}}}\frac{{C}^{{{2}}}}{{N}}\cdot{m}^{{{2}}}\right)}{\left({3}\times{10}^{{{8}}}\frac{{m}}{{s}}\right)}}}}}}\)

\(\displaystyle={1026}\frac{{V}}{{m}}\)

The maximum electric field is related to the maximum magnetic field and the relationship between both of them is given by equation (32.4) in the form

\(\displaystyle{B}_{{\max}}={\frac{{{E}_{{\max}}}}{{{c}}}}\)

Where c is the speed of the light. Now, plug the values for \(\displaystyle{E}_{{\max}}\) and c to get \(\displaystyle{B}_{{\max}}\)

\(\displaystyle{B}_{{\max}}={\frac{{{E}_{{\max}}}}{{{c}}}}={\frac{{{1026}\frac{{V}}{{m}}}}{{{3}\times{10}^{{{8}}}\frac{{m}}{{s}}}}}={3.42}\times{10}^{{-{6}}}{T}\)

\(\displaystyle{I}={\frac{{{1}}}{{{2}}}}\epsilon_{{{o}}}{c}{{E}_{{\max}}^{{{2}}}}\) (1)

Where \(\displaystyle\epsilon_{{{o}}}\) is the electric constant, cc is the speed of light. Solve equation (1) for \(\displaystyle{E}_{{\max}}\)

\(\displaystyle{E}_{{\max}}=\sqrt{{{\frac{{{2}{I}}}{{\epsilon_{{{o}}}{c}}}}}}\) (2)

Plug the values for I, c, and \(\displaystyle\epsilon_{{{o}}}\) into equation (2) to get \(\displaystyle{E}_{{\max}}\)

\(\displaystyle{E}_{{\max}}=\sqrt{{{\frac{{{2}{I}}}{{\epsilon_{{{o}}}{c}}}}}}\)

\(\displaystyle=\sqrt{{{\frac{{{2}{\left({1400}\frac{{W}}{{m}^{{{2}}}}\right)}}}{{{\left({8.85}\times{10}^{{-{12}}}\frac{{C}^{{{2}}}}{{N}}\cdot{m}^{{{2}}}\right)}{\left({3}\times{10}^{{{8}}}\frac{{m}}{{s}}\right)}}}}}}\)

\(\displaystyle={1026}\frac{{V}}{{m}}\)

The maximum electric field is related to the maximum magnetic field and the relationship between both of them is given by equation (32.4) in the form

\(\displaystyle{B}_{{\max}}={\frac{{{E}_{{\max}}}}{{{c}}}}\)

Where c is the speed of the light. Now, plug the values for \(\displaystyle{E}_{{\max}}\) and c to get \(\displaystyle{B}_{{\max}}\)

\(\displaystyle{B}_{{\max}}={\frac{{{E}_{{\max}}}}{{{c}}}}={\frac{{{1026}\frac{{V}}{{m}}}}{{{3}\times{10}^{{{8}}}\frac{{m}}{{s}}}}}={3.42}\times{10}^{{-{6}}}{T}\)