A long straight wire carrying a current of 30 | vedclass

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(C) The external magnetic field is $B_1 = 4 	imes 10^{-4}\,T$,which is parallel to the current. The magnetic field produced by the long straight wire

Vedclass · Accepted Answer

$5 	imes 10^{-4}$

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(C) The external magnetic field is $B_1 = 4 	imes 10^{-4}\,T$,which is parallel to the current. The magnetic field produced by the long straight wire at a distance $r = 2.0\,cm = 2 	imes 10^{-2}\,m$ is given by $B_2 = \frac{\mu_0 I}{2 \pi r} = 2 	imes 10^{-7} 	imes \frac{I}{r}$. Substituting the values,$B_2 = 2 	imes 10^{-7} 	imes \frac{30}{2 	imes 10^{-2}} = 3 	imes 10^{-4}\,T$. Since the field $B_1$ is parallel to the current and the field $B_2$ (due to the wire) is perpendicular to the current (by the right-hand rule),the two fields are mutually perpendicular. The resultant magnetic induction is $B_{	ext{net}} = \sqrt{B_1^2 + B_2^2} = \sqrt{(4 	imes 10^{-4})^2 + (3 	imes 10^{-4})^2} = \sqrt{16 	imes 10^{-8} + 9 	imes 10^{-8}} = \sqrt{25 	imes 10^{-8}} = 5 	imes 10^{-4}\,T$.

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