Answer
$B_m=3.43\mu T$
Work Step by Step
We know that;
$B_m=\frac{E_m}{c}$
We plug in the known values to obtain:
$B_m=\frac{1.03\times 10^3}{3\times 10^8}=3.43\times 10^{-6}=3.43\mu T$
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