A very long wire carrying a current I = 4 sqr | vedclass

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(D) Given: $\mu_0 = 4 \pi 	imes 10^{-7} \, H/m$, distance $d = 20 \, cm = 0.2 \, m$, and current $I = 4 \sqrt{2} \, A$.For a semi-infinite wire, the

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(D) Given: $\mu_0 = 4 \pi 	imes 10^{-7} \, H/m$, distance $d = 20 \, cm = 0.2 \, m$, and current $I = 4 \sqrt{2} \, A$.For a semi-infinite wire, the magnetic field at a perpendicular distance $d$ from one end is given by $B = \frac{\mu_0 I}{4 \pi d}$.For wire $1$, the magnetic field at point $P$ is $B_1 = \frac{\mu_0 I}{4 \pi d} = 10^{-7} 	imes \frac{4 \sqrt{2}}{0.2} = 2 \sqrt{2} 	imes 10^{-6} \, T$.For wire $2$, the magnetic field at point $P$ is $B_2 = \frac{\mu_0 I}{4 \pi d} = 10^{-7} 	imes \frac{4 \sqrt{2}}{0.2} = 2 \sqrt{2} 	imes 10^{-6} \, T$.Since the magnetic fields $B_1$ and $B_2$ are perpendicular to each other, the resultant magnetic field $B$ at point $P$ is:$B = \sqrt{B_1^2 + B_2^2} = \sqrt{(2 \sqrt{2} 	imes 10^{-6})^2 + (2 \sqrt{2} 	imes 10^{-6})^2}$$B = \sqrt{8 	imes 10^{-12} + 8 	imes 10^{-12}} = \sqrt{16 	imes 10^{-12}} = 4 	imes 10^{-6} \, T = 4 \, \mu T$.

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