Calculating the Magnetic Field Due to a Current — Question 5

Question 5

Perpendicular wires produce a vector field that cannot be added as scalars

One infinite wire lies along x with current I toward +x; a second lies along y with the same current toward +y. They cross without electrical contact. Consider $P=(0,0,h)$, with h positive; wire thickness is negligible. Use vacuum permeability $\mu_0\simeq4\pi\times10^{-7}\,\mathrm{T\,m/A}$ where numerical values are needed.

1. Find the total magnetic-field vector and magnitude at P.

2. Choose the magnitude and direction of the y-wire current that makes the total field point along $\hat x-2\hat y$.

Question 5 - Solution

Perpendicular wires produce a vector field that cannot be added as scalars

(a) Method and calculation

$$\vec B_x=-\frac{\mu_0 I}{2\pi h}\hat y,\quad\vec B_y=+\frac{\mu_0 I}{2\pi h}\hat x.$$$$\vec B=\frac{\mu_0 I}{2\pi h}(\hat x-\hat y),\quad|B|=\boxed{\frac{\sqrt2\mu_0 I}{2\pi h}}.$$

(b) Interpretation and further calculation

$$\vec B=\frac{\mu_0}{2\pi h}(I_y\hat x-I\hat y)\parallel(\hat x-2\hat y).$$$$I/I_y=2\Rightarrow\boxed{I_y=I/2\text{ toward }+y}.$$

Physical check

The two fields are perpendicular at P; equal currents give a magnitude larger than either field by a factor of square root of two.