# Magnetic Force on a Current-Carrying Conductor

### Learning Objectives

By the end of this section, you will be able to:- Describe the effects of a magnetic force on a current-carrying conductor.
- Calculate the magnetic force on a current-carrying conductor.

Because charges ordinarily cannot escape a conductor, the magnetic force on charges moving in a conductor is transmitted to the conductor itself.

*v*

_{d}is given by F =

*qv*

_{d}

*B*sin

*θ*. Taking

*B*to be uniform over a length of wire

*l*and zero elsewhere, the total magnetic force on the wire is then F = (

*qv*

_{d}

*B*sin

*θ*)(

*N*), where

*N*is the number of charge carriers in the section of wire of length

*. Now,*

*l**N = 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 = Al*, where

*A*is the cross-sectional area of the wire, then the force on the wire is F = (

*qv*

_{d}

*B*sin

*θ*) (

*nAl*). Gathering terms,

$F=(nqAv_{\text{d}})lB\sin\theta\\$

.
Because *nqAv*

_{d }=

*I*(see Current),

$F=IlB\sin\theta\\$

is the equation for *magnetic force on a length*, as shown in Figure 2. If we divide both sides of this expression by

*l*of wire carrying a current*I*in a uniform magnetic field*B**l*, we find that the magnetic force per unit length of wire in a uniform field is

$\frac{F}{l}=IB\sin\theta\\$

. The direction of this force is given by RHR-1, 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 Current-Carrying Wire: A Strong Magnetic Field

Calculate the force on the wire shown in Figure 1, given*B*= 1.50 T,

*l*= 5.00 cm, and

*I*= 20.0 A.

**Strategy**

The force can be found with the given information by using

$F=IlB\sin\theta\\$

and noting that the angle *θ*between

*I*and

*B*is 90º, so that sin

*θ =*1.

**Solution**

Entering the given values into

*F*=

*IlB*sin

*θ*yields

*F = **IlB* sin *θ* = (20.0 A)(0.0500 m)(1.50 T)(1).

The units for tesla are

This large magnetic field creates a significant force on a small length of wire.

$1\text{ T}=\frac{\text{N}}{\text{A}\cdot\text{m}}\\$

; thus,*F*= 1.50 N.

**Discussion**This large magnetic field creates a significant force on a small length of wire.

## Section Summary

- The magnetic force on current-carrying conductors is given by

where$F=IlB\sin\theta\\$*I**l*is the length of a straight conductor in a uniform magnetic field*B*, and*θ**I*and*B*. The force follows RHR-1 with the thumb in the direction of*I*.

### Conceptual Questions

- Draw a sketch of the situation in Figure 1 showing the direction of electrons carrying the current, and use RHR-1 to verify the direction of the force on the wire.
- Verify 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.
- Why would a magnetohydrodynamic drive work better in ocean water than in fresh water? Also, why would superconducting magnets be desirable?
- Which is more likely to interfere with compass readings, AC current in your refrigerator or DC current when you start your car? Explain.

### Problems & Exercises

1. What is the direction of the magnetic force on the current in each of the six cases in Figure 5?

2. What 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*?
3. What 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

**B**is perpendicular to**I**?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 3.00 × 10

^{−5}-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?
5. (a) A DC power line for a light-rail system carries 1000 A at an angle of 30º to the Earth’s 5.00 × 10

^{−5}-T field. What is the force on a 100-m section of this line? (b) Discuss practical concerns this presents, if any.6. What force is exerted on the water in an MHD drive utilizing a 25.0-cm-diameter tube, if 100-A current is passed across the tube that is perpendicular to a 2.00-T magnetic field? (The relatively small size of this force indicates the need for very large currents and magnetic fields to make practical MHD drives.)

7. A wire carrying a 30.0-A current passes between the poles of a strong magnet that is perpendicular to its field and experiences a 2.16-N force on the 4.00 cm of wire in the field. What is the average field strength?

8. (a) A 0.750-m-long section of cable carrying current to a car starter motor makes an angle of 60º with the Earth’s 5.50 × 10

^{−5 }T field. What is the current when the wire experiences a force of 7.00 × 10^{−3 }N? (b) If you run the wire between the poles of a strong horseshoe magnet, subjecting 5.00 cm of it to a 1.75-T field, what force is exerted on this segment of wire?
9. (a) What is the angle between a wire carrying an 8.00-A current and the 1.20-T 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 90º with the field?

10. The 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 20.0-cm-wide loop?

### Selected Solutions to Problems & Exercises

1. (a) west (left)(b) into page

(c) north (up)

(d) no force

(e) east (right)

(f) south (down)

3. (a) into page

(b) west (left)

(c) out of page

5. (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.

7. 1.80 T

9. (a) 30º (b) 4.80 N

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