Two long parallel wires run along the x axis. Wire 1 is on the x axis (y = 0, z = 0) carrying 28.0 A to the left (−x direction). Wire 2 lies along the line (y = 0.280 m, z = 0) carrying 56.0 A to the right (+x direction). (a) Where in the plane of the two wires is the total magnetic field zero? (b) A particle with charge −2.00 µC moves with velocity 150î Mm/s along the line (y = 0.100 m, z = 0). What is the vector magnetic force on the particle? (c) What vector electric field is required so the particle passes through undeflected?

What you are combining in parts (b) and (c)

Each long straight wire creates a magnetic field of magnitude $$B=\frac{\mu_0 I}{2\pi r}$$ that circles around the wire. Since both wires lie along the $x$ direction, the magnetic field at points in the plane $z=0$ points purely along $\pm\hat{k}$ (into or out of the page). Then the magnetic force is $$\vec F_B=q\,\vec v\times\vec B,$$ and “undeflected” means the net force is zero: $q\vec E+q\vec v\times\vec B=0$.

(b) Net magnetic field at $y=0.100\,\text{m}$

Point: $y=0.100\,\text{m},\ z=0$.

Field from wire 1 (at $y_1=0$) with current to the left ($-\hat{i}$):

• Distance: $r_1=0.100\,\text{m}$
• Magnitude: $$B_1=\frac{\mu_0(28.0)}{2\pi(0.100)}=5.60\times10^{-5}\,\text{T}$$
• Direction: using $\hat{u}\times\hat{\rho}$ with $\hat{u}=-\hat{i}$ and $\hat{\rho}=+\hat{\jmath}$ gives $-\hat{k}$. So, $\vec B_1=-(5.60\times10^{-5})\hat{k}\,\text{T}$.

Field from wire 2 (at $y_2=0.280\,\text{m}$) with current to the right ($+\hat{i}$):

• Distance: $r_2=|0.100-0.280|=0.180\,\text{m}$
• Magnitude: $$B_2=\frac{\mu_0(56.0)}{2\pi(0.180)}=6.22\times10^{-5}\,\text{T}$$
• Direction: here $\hat{\rho}=-\hat{\jmath}$, so $\hat{i}\times(-\hat{\jmath})=-\hat{k}$. So, $\vec B_2=-(6.22\times10^{-5})\hat{k}\,\text{T}$.

Net field: $$\vec B=\vec B_1+\vec B_2=-(1.18\times10^{-4})\hat{k}\,\text{T}.$$

Now compute the force.

• Charge: $q=-2.00\,\mu\text{C}=-2.00\times10^{-6}\,\text{C}$
• Velocity: $150\,\hat{i}\,\text{Mm/s}=150\times10^{6}\,\hat{i}\,\text{m/s}=1.50\times10^{8}\,\hat{i}\,\text{m/s}$

Cross product: $$\vec v\times\vec B=(1.50\times10^{8}\hat{i})\times\left(-(1.18\times10^{-4})\hat{k}\right) =+(1.77\times10^{4})\hat{\jmath}.$$ Then $$\vec F_B=q(\vec v\times\vec B)=(-2.00\times10^{-6})(1.77\times10^{4})\hat{\jmath} =-(3.55\times10^{-2})\hat{\jmath}\,\text{N}.$$

(c) Electric field for zero deflection

“No deflection” means total force is zero: $$q\vec E+q\vec v\times\vec B=0 \quad\Rightarrow\quad \vec E=-\vec v\times\vec B.$$ So, $$\vec E=-(1.77\times10^{4})\hat{\jmath}\,\text{N/C}.$$

Quick direction check (sanity check)

Since $\vec B$ points in $-\hat{k}$ and $\vec v$ is $+\hat{i}$, $\vec v\times\vec B$ points $+\hat{\jmath}$. Because the charge is negative, the magnetic force flips to $-\hat{\jmath}$, and the required electric field must point $-\hat{\jmath}$ so that $q\vec E$ points $+\hat{\jmath}$ and cancels it.