228. Magnetic Force on a CurrentCarrying ConductorLearning Objectives
Because charges ordinarily cannot escape a conductor, the magnetic force on charges moving in a conductor is transmitted to the conductor itself. We can derive an expression for the magnetic force on a current by taking a sum of the magnetic forces on individual charges. (The forces add because they are in the same direction.) The force on an individual charge moving at the drift velocity ${v}_{\mathrm{d}}$ is given by $F={\text{qv}}_{\mathrm{d}}B\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta $. Taking $B$ to be uniform over a length of wire $l$ and zero elsewhere, the total magnetic force on the wire is then $F=({\text{qv}}_{\mathrm{d}}B\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta )(N)$, where $N$ is the number of charge carriers in the section of wire of length $l$. Now, $N=\text{nV}$, where $n$ is the number of charge carriers per unit volume and $V$ is the volume of wire in the field. Noting that $V=\text{Al}$, where $A$ is the crosssectional area of the wire, then the force on the wire is $F=({\text{qv}}_{\mathrm{d}}B\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta )(\text{nAl})$. Gathering terms, $$F=({\text{nqAv}}_{\mathrm{d}})\text{lB}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta .$$Because ${\text{nqAv}}_{\mathrm{d}}=I$ (see Current), $$F=\text{IlB}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta $$is the equation for magnetic force on a length $l$ of wire carrying a current $I$ in a uniform magnetic field $B$, as shown in Figure 2. If we divide both sides of this expression by $l$, we find that the magnetic force per unit length of wire in a uniform field is $\frac{F}{l}=\text{IB}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta $. The direction of this force is given by RHR1, with the thumb in the direction of the current $I$. Then, with the fingers in the direction of $B$, a perpendicular to the palm points in the direction of $F$, as in Figure 2. Example 1: Calculating Magnetic Force on a CurrentCarrying Wire: A Strong Magnetic FieldCalculate the force on the wire shown in Figure 1, given $B=1\text{.}\text{50 T}$, $l=5\text{.}\text{00 cm}$, and $I=\text{20}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{A}$. Strategy The force can be found with the given information by using $F=\text{IlB}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta $ and noting that the angle $\theta $ between $I$ and $B$ is $\text{90\xba}$, so that $\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =1$. Solution Entering the given values into $F=\text{IlB}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta $ yields $$F=\text{IlB}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =\left(\text{20}\text{.0 A}\right)\left(0\text{.}\text{0500 m}\right)\left(1\text{.}\text{50 T}\right)\left(1\right)\text{.}$$The units for tesla are $\text{1 T}=\frac{\mathrm{N}}{\mathrm{A}\cdot \mathrm{m}}$; thus, $$F=1\text{.}\text{50 N.}$$Discussion This large magnetic field creates a significant force on a small length of wire. Magnetic force on currentcarrying conductors is used to convert electric energy to work. (Motors are a prime example—they employ loops of wire and are considered in the next section.) Magnetohydrodynamics (MHD) is the technical name given to a clever application where magnetic force pumps fluids without moving mechanical parts. (See Figure 3.) A strong magnetic field is applied across a tube and a current is passed through the fluid at right angles to the field, resulting in a force on the fluid parallel to the tube axis as shown. The absence of moving parts makes this attractive for moving a hot, chemically active substance, such as the liquid sodium employed in some nuclear reactors. Experimental artificial hearts are testing with this technique for pumping blood, perhaps circumventing the adverse effects of mechanical pumps. (Cell membranes, however, are affected by the large fields needed in MHD, delaying its practical application in humans.) MHD propulsion for nuclear submarines has been proposed, because it could be considerably quieter than conventional propeller drives. The deterrent value of nuclear submarines is based on their ability to hide and survive a first or second nuclear strike. As we slowly disassemble our nuclear weapons arsenals, the submarine branch will be the last to be decommissioned because of this ability (See Figure 4.) Existing MHD drives are heavy and inefficient—much development work is needed. Section Summary
Conceptual QuestionsExercise 1Draw a sketch of the situation in Figure 1 showing the direction of electrons carrying the current, and use RHR1 to verify the direction of the force on the wire. Exercise 2Verify that the direction of the force in an MHD drive, such as that in Figure 3, does not depend on the sign of the charges carrying the current across the fluid. Exercise 3Why would a magnetohydrodynamic drive work better in ocean water than in fresh water? Also, why would superconducting magnets be desirable? Exercise 4Which is more likely to interfere with compass readings, AC current in your refrigerator or DC current when you start your car? Explain. Problems & ExercisesExercise 1What is the direction of the magnetic force on the current in each of the six cases in Figure 5? Figure 5 Show/Hide Solution Solution(a) west (left) (b) into page (c) north (up) (d) no force (e) east (right) (f) south (down) Exercise 2What is the direction of a current that experiences the magnetic force shown in each of the three cases in Figure 6, assuming the current runs perpendicular to $B$? Figure 6 Exercise 3What is the direction of the magnetic field that produces the magnetic force shown on the currents in each of the three cases in Figure 7, assuming $\mathbf{\text{B}}$ is perpendicular to $\mathbf{\text{I}}$? Figure 7 Show/Hide Solution Solution(a) into page (b) west (left) (c) out of page Exercise 4(a) What is the force per meter on a lightning bolt at the equator that carries 20,000 A perpendicular to the Earth’s $3\text{.}\text{00}\times {\text{10}}^{5}\text{T}$ field? (b) What is the direction of the force if the current is straight up and the Earth’s field direction is due north, parallel to the ground? Exercise 5(a) A DC power line for a lightrail system carries 1000 A at an angle of $\text{30.0\xba}$ to the Earth’s $\text{5.00}\times {\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{T}$ field. What is the force on a 100m section of this line? (b) Discuss practical concerns this presents, if any. Show/Hide Solution Solution(a) 2.50 N (b) This is about half a pound of force per 100 m of wire, which is much less than the weight of the wire itself. Therefore, it does not cause any special concerns. Exercise 6What force is exerted on the water in an MHD drive utilizing a 25.0cmdiameter tube, if 100A current is passed across the tube that is perpendicular to a 2.00T magnetic field? (The relatively small size of this force indicates the need for very large currents and magnetic fields to make practical MHD drives.) Exercise 7A wire carrying a 30.0A current passes between the poles of a strong magnet that is perpendicular to its field and experiences a 2.16N force on the 4.00 cm of wire in the field. What is the average field strength? Show/Hide Solution Solution1.80 T Exercise 8(a) A 0.750mlong section of cable carrying current to a car starter motor makes an angle of $\text{60\xba}$ with the Earth’s $5\text{.}\text{50}\times {\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{T}$ field. What is the current when the wire experiences a force of $\text{7.00}\times {\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\mathrm{N}$? (b) If you run the wire between the poles of a strong horseshoe magnet, subjecting 5.00 cm of it to a 1.75T field, what force is exerted on this segment of wire? Exercise 9(a) What is the angle between a wire carrying an 8.00A current and the 1.20T field it is in if 50.0 cm of the wire experiences a magnetic force of 2.40 N? (b) What is the force on the wire if it is rotated to make an angle of $\text{90\xba}$ with the field? Show/Hide Solution Solution(a) $\text{30\xba}$ (b) 4.80 N Exercise 10The force on the rectangular loop of wire in the magnetic field in Figure 8 can be used to measure field strength. The field is uniform, and the plane of the loop is perpendicular to the field. (a) What is the direction of the magnetic force on the loop? Justify the claim that the forces on the sides of the loop are equal and opposite, independent of how much of the loop is in the field and do not affect the net force on the loop. (b) If a current of 5.00 A is used, what is the force per tesla on the 20.0cmwide loop?
