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Two long, straight wires are oriented perpendicular to the
Chapter 22, Problem 54P(choose chapter or problem)
Two long, straight wires are oriented perpendicular to the page, as shown in Figure 22–40. The current in one wire is \I_{1}=3.0 \mathrm{~A}\), pointing into the page, and the current in the other wire is \(I_{2}=4.0 \mathrm{~A}\), pointing out of the page. Find the magnitude and direction of the net magnetic field at point P.
Questions & Answers
QUESTION:
Two long, straight wires are oriented perpendicular to the page, as shown in Figure 22–40. The current in one wire is \I_{1}=3.0 \mathrm{~A}\), pointing into the page, and the current in the other wire is \(I_{2}=4.0 \mathrm{~A}\), pointing out of the page. Find the magnitude and direction of the net magnetic field at point P.
Step 1 of 2
Here, we have to find the magnitude and the direction of the magnetic field at point P.
Given that,
\(\begin{array}{l} I_{1}=3 \mathrm{~A} . \\ I_{2}=4 \mathrm{~A} . \end{array}\)
The net magnetic field must be the resultant of the two magnetic fields by the two currents.
So,
\(\begin{aligned} B_{1} & =\frac{\mu_{0} I_{1}}{2 \pi r_{1}} \widehat{x} . \\ B_{2} & =\frac{\mu_{0} I_{2}}{2 \pi r}\left(-\cos 45^{0} \widehat{x}+\sin 45^{0} \widehat{y}\right) \\ & =\frac{\mu_{0} I_{2}}{2 \pi r_{2}}(-0.707 \widehat{x}+0.707 \widehat{y}) \\ & =\frac{\mu_{0} I_{2}}{2 \pi r_{2}} \times 0.707(-\widehat{x}+\widehat{y}) . \end{aligned}\)