Chapter 20, Problem 50P

Set Up: $B=\frac{{\mu }_{0}l}{2\pi r}$ The direction of $\stackrel{\to }{B}$B is given by the right-hand rule in Section 20.7.

Solve: (a) The magnetic fields of the two wires are in opposite directions at points between the wires, as shown in Figure (a) below. Consider a point that is a distance x from wire 1 and hence 0.400 m − x from wire 2. B₁ = B₂ gives

$\frac{{\mu }_0{I}_1}{2\pi x}=\frac{{\mu }_0{I}_2}{2\pi \left(0.400\text{m}-x\right)}.\frac{25.0\text{A}}{x}=\frac{75.0\text{A}}{\left(0.400\text{m}-x\right)}\mathrm{.3.00}x=0.400\text{m}-x$ and x = 0.100 m

The net field is zero at a point between the wires, 10.0 cm from the wire carrying 25.0 A and 30.0 cm from the wire carrying 75.0 A.

50. Two long, parallel transmission lines 40.0 cm apart carry 25.0 A and 75.0 A currents. Find all locations where the net magnetic field of the two wires is zero if these currents are in (a) the same direction, (b) opposite directions.