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what should the direction of the magnetic field be in order to produce the force in part (b)?

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Magnetic Forces and Fields

72 Magnetic Force on a Current-Carrying Usher

Learning Objectives

Past the end of this section, you will be able to:

  • Determine the direction in which a electric current-carrying wire experiences a forcefulness in an external magnetic field
  • Calculate the force on a electric current-carrying wire in an external magnetic field

Moving charges experience a force in a magnetic field. If these moving charges are in a wire—that is, if the wire is carrying a current—the wire should besides feel a force. However, earlier we hash out the force exerted on a current by a magnetic field, we commencement examine the magnetic field generated by an current. We are studying two separate effects here that interact closely: A current-carrying wire generates a magnetic field and the magnetic field exerts a force on the current-carrying wire.

Magnetic Fields Produced by Electric Currents

When discussing historical discoveries in magnetism, nosotros mentioned Oersted'south finding that a wire carrying an electrical current acquired a nearby compass to deflect. A connectedness was established that electrical currents produce magnetic fields. (This connectedness betwixt electricity and magnetism is discussed in more detail in Sources of Magnetic Fields.)

The compass needle nigh the wire experiences a force that aligns the needle tangent to a circle around the wire. Therefore, a electric current-conveying wire produces circular loops of magnetic field. To determine the direction of the magnetic field generated from a wire, nosotros apply a 2nd right-paw rule. In RHR-two, your pollex points in the direction of the electric current while your fingers wrap around the wire, pointing in the direction of the magnetic field produced ((Figure)). If the magnetic field were coming at you lot or out of the folio, we represent this with a dot. If the magnetic field were going into the folio, we represent this with an ×. These symbols come from because a vector arrow: An arrow pointed toward yous, from your perspective, would look similar a dot or the tip of an pointer. An arrow pointed abroad from you, from your perspective, would look like a cross or an ×. A composite sketch of the magnetic circles is shown in (Figure), where the field strength is shown to decrease every bit yous get farther from the wire by loops that are farther separated.

(a) When the wire is in the aeroplane of the newspaper, the field is perpendicular to the paper. Notation the symbols used for the field pointing inward (like the tail of an arrow) and the field pointing outward (similar the tip of an arrow). (b) A long and straight wire creates a field with magnetic field lines forming circular loops.

Figure a shows the right hand rule applied to a wire carrying an upward current. The right hand is placed so that the thumb points up, in the direction of the current. The fingers curl around the wire. The magnetic field is out of the page to the left of the current and into the page to the right of the current. Figure b shows the magnetic field lines for the upward current. The field lines form concentric rings that circulate in the same direction as the fingers of the right hand in figure a. The spacing between rings increases with distance from the current.

Computing the Magnetic Strength

Electric current is an ordered movement of charge. A current-conveying wire in a magnetic field must therefore experience a force due to the field. To investigate this force, let'south consider the minute section of wire every bit shown in (Figure). The length and cross-sectional expanse of the section are dl and A, respectively, so its volume is V=A·dl. The wire is formed from cloth that contains n accuse carriers per unit book, so the number of charge carriers in the department is nA·dl. If the charge carriers motion with drift velocity {\stackrel{\to }{v}}_{\text{d}}, the electric current I in the wire is (from Current and Resistance)

I=neA{v}_{d}.

The magnetic force on any single charge carrier is e{\stackrel{\to }{v}}_{\text{d}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{B}, so the full magnetic force d\stackrel{\to }{F} on the nA·dl charge carriers in the section of wire is

d\stackrel{\to }{F}=\left(nA·dl\right)e{\stackrel{\to }{v}}_{\text{d}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{B}.

Nosotros can define dl to be a vector of length dl pointing along {\stackrel{\to }{v}}_{\text{d}}, which allows us to rewrite this equation every bit

d\stackrel{\to }{F}=neA{v}_{\text{d}}\stackrel{\to }{dl}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{B},

or

d\stackrel{\to }{F}=I\stackrel{\to }{dl}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{B}.

This is the magnetic force on the section of wire. Annotation that information technology is really the internet strength exerted by the field on the charge carriers themselves. The management of this force is given past RHR-ane, where you bespeak your fingers in the management of the current and gyre them toward the field. Your pollex so points in the direction of the force.

An infinitesimal section of current-conveying wire in a magnetic field.

An illustration of a curving current-carrying wire in a uniform magnetic field. A detail view of a small segment of the wire shows a short, straight piece of current, length d l with current I though it. The velocity v sub d is in the direction of the current. The field B makes an angle theta with the velocity vector.

To make up one's mind the magnetic force \stackrel{\to }{F} on a wire of arbitrary length and shape, we must integrate (Figure) over the entire wire. If the wire section happens to be straight and B is compatible, the equation differentials go accented quantities, giving us

\stackrel{\to }{F}=I\stackrel{\to }{l}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{B}.

This is the strength on a straight, current-carrying wire in a uniform magnetic field.

Balancing the Gravitational and Magnetic Forces on a Electric current-Carrying Wire A wire of length fifty cm and mass 10 thou is suspended in a horizontal aeroplane by a pair of flexible leads ((Figure)). The wire is and then subjected to a abiding magnetic field of magnitude 0.50 T, which is directed as shown. What are the magnitude and direction of the current in the wire needed to remove the tension in the supporting leads?

(a) A wire suspended in a magnetic field. (b) The free-torso diagram for the wire.

Figure a: An illustration of a wire suspended in a uniform magnetic field. The magnetic field points into the page. The wire is horizontal and is 50 cm long. A variable voltage source completes the circuit made by the wire and the leads used to suspend it. Figure b: A free body diagram of the wire. The current is to the right. The weight, mg, points down. Tension at either end is up. The magnetic force is up.

Strategy From the costless-torso diagram in the effigy, the tensions in the supporting leads go to zero when the gravitational and magnetic forces balance each other. Using the RHR-1, we find that the magnetic force points up. We can then determine the electric current I by equating the 2 forces.

Solution Equate the two forces of weight and magnetic force on the wire:

mg=IlB.

Thus,

I=\frac{mg}{lB}=\frac{\left(0.010\phantom{\rule{0.2em}{0ex}}\text{kg}\right)\left(9.8{\phantom{\rule{0.2em}{0ex}}\text{m/s}}^{2}\right)}{\left(0.50\phantom{\rule{0.2em}{0ex}}\text{m}\right)\left(0.50\phantom{\rule{0.2em}{0ex}}\text{T}\right)}=0.39\phantom{\rule{0.2em}{0ex}}\text{A.}

Significance This big magnetic field creates a significant force on a length of wire to counteract the weight of the wire.

Calculating Magnetic Force on a Current-Carrying Wire A long, rigid wire lying along the y-centrality carries a 5.0-A electric current flowing in the positive y-direction. (a) If a abiding magnetic field of magnitude 0.xxx T is directed along the positive x-axis, what is the magnetic force per unit length on the wire? (b) If a abiding magnetic field of 0.30 T is directed 30 degrees from the +10-axis towards the +y-axis, what is the magnetic force per unit of measurement length on the wire?

Strategy The magnetic force on a current-carrying wire in a magnetic field is given past \stackrel{\to }{F}=I\stackrel{\to }{l}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{B}. For part a, since the electric current and magnetic field are perpendicular in this problem, we can simplify the formula to give us the magnitude and discover the direction through the RHR-1. The angle θ is 90 degrees, which means \text{sin}\phantom{\rule{0.1em}{0ex}}\theta =1. Besides, the length can be divided over to the left-paw side to discover the force per unit length. For part b, the current times length is written in unit of measurement vector notation, likewise equally the magnetic field. Afterward the cantankerous product is taken, the directionality is evident by the resulting unit vector.

Solution

  1. We first with the general formula for the magnetic force on a wire. Nosotros are looking for the force per unit of measurement length, so nosotros divide past the length to bring it to the left-hand side. We also prepare \text{sin}\phantom{\rule{0.1em}{0ex}}\theta =1. The solution therefore is

    \begin{array}{ccc}\hfill F& =\hfill & IlB\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}\theta \hfill \\ \hfill \frac{F}{l}& =\hfill & \left(5.0\phantom{\rule{0.2em}{0ex}}\text{A}\right)\left(\text{0.30 T}\right)\hfill \\ \hfill \frac{F}{l}& =\hfill & 1.5\phantom{\rule{0.2em}{0ex}}\text{N/m.}\hfill \end{array}


    Directionality: Point your fingers in the positive y-direction and curl your fingers in the positive x-direction. Your thumb will point in the \text{−}\stackrel{\to }{k} direction. Therefore, with directionality, the solution is

    \frac{\stackrel{\to }{F}}{l}=-1.5\stackrel{\to }{k}\phantom{\rule{0.2em}{0ex}}\text{N/m.}

  2. The electric current times length and the magnetic field are written in unit vector notation. Then, we have the cross product to discover the force:

    \begin{array}{ccc}\hfill \stackrel{\to }{F}& =\hfill & I\stackrel{\to }{l}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{B}=\left(5.0A\right)l\stackrel{^}{j}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(0.30T\text{cos}\left(30\text{°}\right)\stackrel{^}{i}+0.30T\phantom{\rule{0.1em}{0ex}}\text{sin}\left(30\text{°}\right)\stackrel{^}{j}\right)\hfill \\ \hfill \stackrel{\to }{F}\text{/}l& =\hfill & -1.30\stackrel{^}{k}\phantom{\rule{0.2em}{0ex}}\text{N/m.}\hfill \end{array}

Significance This big magnetic field creates a significant force on a small length of wire. As the angle of the magnetic field becomes more closely aligned to the current in the wire, there is less of a strength on it, as seen from comparing parts a and b.

Check Your Understanding A direct, flexible length of copper wire is immersed in a magnetic field that is directed into the page. (a) If the wire's current runs in the +x-management, which way will the wire bend? (b) Which way volition the wire bend if the electric current runs in the –x-direction?

a. bends upward; b. bends downwardly

Force on a Circular Wire A circular current loop of radius R carrying a current I is placed in the xy-plane. A constant uniform magnetic field cuts through the loop parallel to the y-axis ((Effigy)). Find the magnetic force on the upper half of the loop, the lower half of the loop, and the total force on the loop.

A loop of wire conveying a current in a magnetic field.

A loop of radius R is in the plane of the page. The loop carries a clockwise current I and is in a uniform magnetic field that points up the page.

Strategy The magnetic force on the upper loop should be written in terms of the differential strength acting on each segment of the loop. If we integrate over each differential piece, we solve for the overall force on that section of the loop. The force on the lower loop is plant in a similar way, and the total forcefulness is the addition of these two forces.

Solution A differential strength on an arbitrary piece of wire located on the upper ring is:

dF=IB\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}\theta \phantom{\rule{0.1em}{0ex}}dl.

where \theta is the angle between the magnetic field direction (+y) and the segment of wire. A differential segment is located at the same radius, then using an arc-length formula, we take:

\begin{array}{ccc}\hfill dl& =\hfill & R\phantom{\rule{0.1em}{0ex}}d\theta \hfill \\ \hfill dF& =\hfill & IBR\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}\theta \phantom{\rule{0.1em}{0ex}}d\theta .\hfill \end{array}

In society to notice the force on a segment, we integrate over the upper half of the circle, from 0 to \pi . This results in:

F=IBR{\underset{0}{\overset{\pi }{\int }}\text{sin}\phantom{\rule{0.1em}{0ex}}\theta \phantom{\rule{0.1em}{0ex}}d\theta =IBR\left(\text{−}\text{cos}\pi +\text{cos}0\right)}^{}=2IBR.

The lower one-half of the loop is integrated from \pi to zero, giving united states of america:

F=IBR{\underset{\pi }{\overset{0}{\int }}\text{sin}\phantom{\rule{0.1em}{0ex}}\theta \phantom{\rule{0.1em}{0ex}}d\theta =IBR\left(\text{−}\text{cos}0+\text{cos}\pi \right)}^{}=-2IBR.

The net force is the sum of these forces, which is zero.

Significance The total forcefulness on whatsoever airtight loop in a uniform magnetic field is zero. Even though each piece of the loop has a force acting on information technology, the internet force on the arrangement is nada. (Notation that there is a cyberspace torque on the loop, which we consider in the next section.)

Summary

  • An electrical current produces a magnetic field around the wire.
  • The directionality of the magnetic field produced is determined by the correct hand dominion-2, where your thumb points in the direction of the electric current and your fingers wrap effectually the wire in the direction of the magnetic field.
  • The magnetic force on current-carrying conductors is given by \stackrel{\to }{F}=I\stackrel{\to }{l}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{B} where I is the current and l is the length of a wire in a uniform magnetic field B.

Conceptual Questions

Draw the error that results from accidently using your left rather than your right hand when determining the management of a magnetic force.

Because the magnetic force law, are the velocity and magnetic field ever perpendicular? Are the forcefulness and velocity always perpendicular? What about the strength and magnetic field?

Velocity and magnetic field can be set together in whatever direction. If there is a force, the velocity is perpendicular to it. The magnetic field is also perpendicular to the force if information technology exists.

Why tin can a nearby magnet distort a cathode ray tube television film?

A magnetic field exerts a force on the moving electrons in a current carrying wire. What exerts the forcefulness on a wire?

A force on a wire is exerted by an external magnetic field created past a wire or another magnet.

There are regions where the magnetic field of earth is almost perpendicular to the surface of Earth. What difficulty does this cause in the use of a compass?

Bug

What is the direction of the magnetic force on the current in each of the half-dozen cases?

Case a: I is down, B is out of the page. Case b: I is up, B is to the right. Case c: I is to the right, B is into the page. Case d: I is to the left , B is to the right. Case e: I is into the page, B is up. Case f: I is out of the page, B is to the left.

a. left; b. into the page; c. up; d. no strength; e. right; f. downward

What is the management of a current that experiences the magnetic force shown in each of the 3 cases, assuming the current runs perpendicular to \stackrel{\to }{B}?

Case a: B is out of the page, F is up. Case b: B is to the right, F is up. Case c: B is into the page, F is to the left.

What is the direction of the magnetic field that produces the magnetic force shown on the currents in each of the three cases, assuming \stackrel{\to }{B} is perpendicular to I?

Case a: I is up, F is to the left. Case b: I is down, F is into the page. Case c: I is to the left, F is up.

a. into the page; b. left; c. out of the page

(a) What is the force per meter on a lightning bolt at the equator that carries twenty,000 A perpendicular to Earth's 3.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}\text{T} field? (b) What is the management of the force if the current is straight up and Earth'due south field direction is due north, parallel to the footing?

(a) A dc ability line for a lite-rail system carries g A at an angle of 30.0º to Earth'south 5.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}\text{T} field. What is the strength on a 100-m department of this line? (b) Discuss applied concerns this presents, if any.

a. 2.l N; b. This means that the lite-runway power lines must be attached in order not to be moved past the force caused by Globe's magnetic field.

A wire carrying a 30.0-A current passes betwixt the poles of a strong magnet that is perpendicular to its field and experiences a 2.xvi-Due north strength on the 4.00 cm of wire in the field. What is the boilerplate field force?

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Source: https://opentextbc.ca/universityphysicsv2openstax/chapter/magnetic-force-on-a-current-carrying-conductor/

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