Two long parallel wires carry currents of 20 A and 5.0 A in opposite directions.The wires are separated by 20 cm.At what point between the two wires do they produce the same magnitude $\vec { B }$ field?

Question

Quizplus · Accepted Answer

To find the point where the magnetic field of the two wires are equal in magnitude, we can use the equation:

\begin{align*}
B &= \frac{\mu_0}{4\pi}\frac{2I}{d}\
\end{align*}

where $B$ is the magnetic field, $\mu_0$ is the permeability of free space, $I$ is the current, and $d$ is the distance from the wire.

Using this equation, we can find the magnetic field produced by each wire at different distances from the wires:

\begin{align*}
B_{20A} &= \frac{\mu_0}{4\pi}\frac{2(20 	ext{ A})}{0.2 	ext{ m}} = 2.0 	imes 10^{-4} 	ext{ T}\
B_{5A} &= \frac{\mu_0}{4\pi}\frac{2(5 	ext{ A})}{0.2 	ext{ m}} = 5.0 	imes 10^{-5} 	ext{ T}
\end{align*}

At a distance of 0 cm (right on top of the 20 A wire), the magnetic field produced by the 20 A wire is much stronger than the field produced by the 5 A wire. As we move away from the 20 A wire, the magnetic field it produces decreases, while the magnetic field produced by the 5 A wire increases. At some point, the two fields will be equal in magnitude.

To find that point, we can set the two magnetic fields equal to each other and solve for the distance from the 20 A wire:

\begin{align*}
B_{20A} &= B_{5A}\
\frac{\mu_0}{4\pi}\frac{2(20 	ext{ A})}{x} &= \frac{\mu_0}{4\pi}\frac{2(5 	ext{ A})}{0.2 	ext{ m} - x}\
\frac{1}{x} &= \frac{1}{0.2 	ext{ m} - x}	imes \frac{5}{20}\
x &= 0.16 	ext{ m} = 16 	ext{ cm}
\end{align*}

Therefore, the point where the magnetic fields of the two wires are equal in magnitude is 16 cm from the 20 A wire. The answer is D.

Two Long Parallel Wires Carry Currents of 20 a and 5.0