A long straight wire carrying a current of 30 | vedclass

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(C) Given: Current $i = 30 ~A$,External magnetic field $B_1 = 4 	imes 10^{-4} ~T$,Distance $r = 2.0 ~cm = 2 	imes 10^{-2} ~m$.The magnetic field pro

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$5 	imes 10^{-4}$

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(C) Given: Current $i = 30 ~A$,External magnetic field $B_1 = 4 	imes 10^{-4} ~T$,Distance $r = 2.0 ~cm = 2 	imes 10^{-2} ~m$.The magnetic field produced by the straight wire at a distance $r$ is given by $B_2 = \frac{\mu_0 i}{2 \pi r}$.Substituting the values: $B_2 = \frac{2 	imes 10^{-7} 	imes 30}{2 	imes 10^{-2}} = 3 	imes 10^{-4} ~T$.Since the external magnetic field $B_1$ is parallel to the current,it is perpendicular to the magnetic field $B_2$ produced by the wire (which circles the wire).The resultant magnetic field $B$ is given by $B = \sqrt{B_1^2 + B_2^2}$.$B = \sqrt{(4 	imes 10^{-4})^2 + (3 	imes 10^{-4})^2} = \sqrt{16 	imes 10^{-8} + 9 	imes 10^{-8}} = \sqrt{25 	imes 10^{-8}} = 5 	imes 10^{-4} ~T$.

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