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વિદ્યુત ચુંબકીય તરંગમાં વિદ્યુતક્ષેત $\overrightarrow{\mathrm{E}}=\hat{i} 40 \cos \omega(\mathrm{t}-z / \mathrm{c})$ થી આપવામાં આવે છે. આ તરંગમાં ચુંબકીય ક્ષેત્ર. . . . . . . થશે.
$\overrightarrow{\mathrm{B}}=\hat{\mathrm{i}} \frac{40}{\mathrm{c}} \cos \omega\left(\mathrm{t}-\frac{\mathrm{z}}{\mathrm{c}}\right)$
$\vec{B}=\hat{j} 40 \cos \omega\left(t-\frac{z}{c}\right)$
$\overrightarrow{\mathrm{B}}=\hat{\mathrm{k}} \frac{40}{\mathrm{c}} \cos \omega\left(\mathrm{t}-\frac{\mathrm{z}}{\mathrm{c}}\right)$
$\vec{B}=\hat{j} \frac{40}{c} \cos \omega\left(t-\frac{z}{c}\right)$
Solution
$\overrightarrow{\mathrm{E}}=\hat{\mathrm{i}} 40 \cos \omega\left(\mathrm{t}-\frac{\mathrm{z}}{\mathrm{c}}\right)$
$\overrightarrow{\mathrm{E}}$ is along $+\mathrm{x}$ direction
$\overrightarrow{\mathrm{v}}$ is along $+\mathrm{z}$ direction
So direction of $\vec{B}$ will be along +y and magnitude of $B$ will be $\frac{E}{C}$
So answer is $\frac{40}{c} \cos \omega\left(t-\frac{z}{c}\right) \hat{j}$
