Students will learn to analyze and solve problems involving current carrying wires in magnetic fields.

### Key Equations

\begin{align*} F_{\text{wire}}= LIB\sin(\theta) &&\text{Force on a Current Carrying Wire}\end{align*}

In this equation, *L* refers to the length of the wire, *I* to the electric current, *B* the magnitude of the magnetic field and \begin{align*} \theta \end{align*} is the angle between the direction of the current and the direction of the magnetic field.

\begin{align*} B_{\text{wire}} = \frac{\mu_0 I}{2 \pi r}&& \text{Magnetic field at a distance } r \text{ from a current-carrying wire}\\ \text{Where } \mu_0=4 \pi \times 10^{-7} \;\text{Tm/A} &&\text{Permeability of Vacuum (approximately same for air also)} \end{align*}

**Force on a Wire**

Since a wire is nothing but a collection of moving charges, the force it will experience in a magnetic field will simply be the vector sum of the forces on the individual charges. If the wire is straight --- that is, all the charges are moving in the same direction --- these forces will all point in the same direction, and so will their sum. Then, the direction of the force can be found using the second right hand rule, while its magnitude will depend on the length of the wire (denoted \begin{align*} L\end{align*}), the strength of the current, the strength of the field, and the angle between their directions:

Two current-carrying wires next to each other each generate magnetic fields and therefore exert forces on each other:

#### Example 1

#### Example 2

A wire loop and an infinitely long current carrying cable are placed a distance \begin{align*}r\end{align*} apart. The infinitely long wire is carrying a current \begin{align*}I_1\end{align*} to the left and the loop is carrying a current \begin{align*}I_2\end{align*} CCW. The dimensions of the wire loop are shown in the diagram illustrating the situation below. What is the magnitude and direction of the net force on the loop (the mass of the wires are negligible)?

##### Solution

In this problem, it is best to start by determining the direction of the force on each segment of the loop. Based on the first right hand rule, the magnetic field from the infinite cable points into the page where the loop is. This means that the force on the top segment of the loop will be down toward the bottom of the page, the force on the left segment will be to right, the force on the bottom segment will be toward the top of the page, and the force on the right segment will be to the left. The forces on the left and right segments will balance out because both segments are the same distance from the cable. The forces from the top and bottom section will not balance out because the wires are different distances from the cable. The force on the bottom segment will be stronger than the one on the top segment because the magnetic field is stronger closer to the cable, so the net force on the loop will be up, toward the top of the page.

Now we will begin to calculate the force's magnitude by first determining the strength of the magnetic field at the bottom and top segments. All we really have to do is plug in the distances to each segment into the equation we already know for the magnetic field due to a current carrying wire.

\begin{align*} B&=\frac{\mu_oI}{2\pi r}\\ B_{bottom}&=\frac{\mu_oI_1}{2\pi R}\\ B_{top}&=\frac{\mu_oI_1}{2\pi 2R}\\ \end{align*}

Now we will calculate the net force on the loop using the equation given above. We'll consider up the positive direction.

\begin{align*} \Sigma F&=F_{bottom} - F_{top} && \text{start by summing the forces on the loop}\\ \Sigma F&=I_2LB_{bottom} - I_2LB_{top} && \text{substitute in the values for each of the force terms}\\ \Sigma F&=I_2L(B_{bottom} - B_{top}) && \text{factor the equation}\\ \Sigma F&=I_2L(\frac{\mu_oI_1}{2\pi R} - \frac{\mu_oI_1}{2\pi 2R}) && \text{substitute in the values for the magnetic field}\\ \Sigma F&=\frac{\mu_oI_1I_2L}{2\pi R}(1-\frac{1}{2}) && \text{factor the equation again}\\ \Sigma F&=\frac{\mu_oI_1I_2L}{4\pi R} && \text{simplify to get the answer}\\ \end{align*}

### Watch this Explanation

### Explore More

- A vertical wire, with a current of \begin{align*}6.0 \;\mathrm{A}\end{align*} going towards the ground, is immersed in a magnetic field of \begin{align*}5.0 \;\mathrm{T}\end{align*} pointing to the right. What is the value and direction of the force on the wire? The length of the wire is \begin{align*}2.0 \;\mathrm{m}\end{align*}.

- A futuristic magneto-car uses the interaction between current flowing across the magneto car and magnetic fields to propel itself forward. The device consists of two fixed metal tracks and a freely moving metal car (see illustration above). A magnetic field is pointing downward with respect to the car, and has the strength of \begin{align*}5.00 \;\mathrm{T}\end{align*}. The car is \begin{align*}4.70 \;\mathrm{m}\end{align*} wide and has \begin{align*} 800 \;\mathrm{A}\end{align*}of current flowing through it. The arrows indicate the direction of the current flow.
- Find the direction and magnitude of the force on the car.
- If the car has a mass of \begin{align*}2050 \;\mathrm{kg}\end{align*}, what is its velocity after \begin{align*}10 \;\mathrm{s}\end{align*}, assuming it starts at rest?
- If you want double the force for the same magnetic field, how should the current change?

- A horizontal wire carries a current of \begin{align*}48\;\mathrm{A}\end{align*} towards the east. A second wire with mass \begin{align*}0.05 \;\mathrm{kg}\end{align*} runs parallel to the first, but lies \begin{align*}15 \;\mathrm{cm}\end{align*} below it. This second wire is held in suspension by the magnetic field of the first wire above it. If each wire has a length of half a meter, what is the magnitude and direction of the current in the lower wire?
- Show that the formula for the force between two current carrying wires is \begin{align*} F = \frac {\mu_{o} L i_1 i_2} {2 \pi d} \end{align*} , where
*d*is the distance between the two wires, \begin{align*} i_1 \end{align*} is the current of first wire and*L*is the segment of length of the second wire carrying a current \begin{align*} i_2 \end{align*}. (Hint: find magnetic field emanating from first wire and then use the formula for a wire immersed in that magnetic field in order to find the force on the second wire.) - Two long thin wires are on the same plane but perpendicular to each other. The wire on the \begin{align*}y-\end{align*}axis carries a current of \begin{align*}6.0 \;\mathrm{A}\end{align*} in the \begin{align*}-y\end{align*} direction. The wire on the \begin{align*}x-\end{align*}axis carries a current of \begin{align*}2.0 \;\mathrm{A}\end{align*} in the \begin{align*}+ x\end{align*} direction. Point, \begin{align*}P\end{align*} has the co-ordinates of \begin{align*}(2.0, 2,0)\end{align*} in meters. A charged particle moves in a direction of \begin{align*}45^o\end{align*} away from the origin at point, \begin{align*}P\end{align*}, with a velocity of \begin{align*}1.0 \times 10^7 \;\mathrm{m/s}.
\end{align*}
- Find the magnitude and direction of the magnetic field at point, \begin{align*}P\end{align*}.
- If there is a magnetic force of \begin{align*}1.0 \times 10^{-6} \;\mathrm{N}\end{align*} on the particle determine its charge.
- Determine the magnitude of an electric field that will cancel the magnetic force on the particle.

- A long straight wire is on the \begin{align*}x-\end{align*}axis and has a current of \begin{align*}12 \;\mathrm{A}\end{align*} in the \begin{align*}-x\end{align*} direction. A point \begin{align*}P\end{align*}, is located \begin{align*}2.0 \;\mathrm{m}\end{align*} above the wire on the \begin{align*}y-\end{align*}axis.
- What is the magnitude and direction of the magnetic field at \begin{align*}P\end{align*}.
- If an electron moves through \begin{align*}P\end{align*} in the\begin{align*}-x\end{align*} direction at a speed of \begin{align*}8.0 \times 10^7 \;\mathrm{m/s}\end{align*} what is the magnitude and direction of the force on the electron?

#### Answers to Selected Problems

- Down the page; \begin{align*}60 \;\mathrm{N}\end{align*}
- a. To the right, \begin{align*}1.88 \times 10^4 \;\mathrm{N}\end{align*} b. \begin{align*}91.7 \;\mathrm{m/s}\end{align*} c. It should be doubled
- East \begin{align*}1.5 \times 10^4 \;\mathrm{A}\end{align*}
- .
- a. \begin{align*}8 \times 10^{-7} \;\mathrm{T}\end{align*} b. \begin{align*}1.3 \times 10^{-6} \;\mathrm{C}\end{align*}
- a. \begin{align*}1.2 \times 10^{-6} \;\mathrm{T}, +z\end{align*} b. \begin{align*}1.5 \times 10^{-17} \;\mathrm{N}, -y\end{align*}