<< Chapter < Page Chapter >> Page >
  • 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.

A diagram showing a circuit with current I running through it. One section of the wire passes between the north and south poles of a magnet with a diameter l. Magnetic field B is oriented toward the right, from the north to the south pole of the magnet, across the wire. The current runs out of the page. The force on the wire is directed up. An illustration of the right hand rule 1 shows the thumb pointing out of the page in the direction of the current, the fingers pointing right in the direction of B, and the F vector pointing up and away from the palm.
The magnetic field exerts a force on a current-carrying wire in a direction given by the right hand rule 1 (the same direction as that on the individual moving charges). This force can easily be large enough to move the wire, since typical currents consist of very large numbers of moving charges.

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 d is given by F = qv d B sin θ . Taking B size 12{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 ) size 12{F= \( ital "qv" rSub { size 8{d} } B"sin"θ \) \( N \) } {} , where N size 12{N} {} is the number of charge carriers in the section of wire of length l size 12{l} {} . Now, N = nV size 12{N= ital "nV"} {} , where n size 12{n} {} is the number of charge carriers per unit volume and V size 12{V} {} is the volume of wire in the field. Noting that V = Al size 12{V= ital "Al"} {} , where A size 12{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 d ) lB sin θ . size 12{F= \( ital "nqAv" rSub { size 8{d} } \) ital "lB""sin"θ} {}

Because nqAv d = I size 12{ ital "nqAv" rSub { size 8{d} } =I} {} (see Current ),

F = IlB sin θ size 12{F= ital "IlB""sin"θ} {}

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 [link] . 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 F l = IB sin θ size 12{ { {F} over {l} } = ital "IB""sin"θ} {} . The direction of this force is given by RHR-1, with the thumb in the direction of the current I size 12{I} {} . Then, with the fingers in the direction of B size 12{B} {} , a perpendicular to the palm points in the direction of F size 12{F} {} , as in [link] .

Illustration of the right hand rule 1 showing the thumb pointing right in the direction of current I, the fingers pointing into the page with magnetic field B, and the force directed up, away from the palm.
The force on a current-carrying wire in a magnetic field is F = IlB sin θ size 12{F= ital "IlB""sin"θ} {} . Its direction is given by RHR-1.

Calculating magnetic force on a current-carrying wire: a strong magnetic field

Calculate the force on the wire shown in [link] , given B = 1 . 50 T size 12{B=1 "." "50"" T"} {} , l = 5 . 00 cm size 12{l=5 "." "00"" cm"} {} , and I = 20 . 0 A size 12{I="20" "." 0 A} {} .

Strategy

The force can be found with the given information by using F = IlB sin θ size 12{F= ital "IlB""sin"θ} {} and noting that the angle θ size 12{θ} {} between I size 12{I} {} and B size 12{B} {} is 90º , so that sin θ = 1 .

Solution

Entering the given values into F = IlB sin θ size 12{F= ital "IlB""sin"θ} {} yields

F = IlB sin θ = 20 .0 A 0 . 0500 m 1 . 50 T 1 . size 12{F= ital "IlB""sin"θ= left ("20" "." 0" A" right ) left (0 "." "0500"" m" right ) left (1 "." "50"" T" right ) left (1 right )} {}

The units for tesla are 1 T = N A m size 12{"1 T"= { {N} over {A cdot m} } } {} ; thus,

F = 1 . 50 N. size 12{F=1 "." "50"" N"} {}

Discussion

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

Magnetic force on current-carrying 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 [link] .)

Diagram showing a cylinder of fluid of diameter l placed between the north and south poles of a magnet. The north pole is to the left. The south pole is to the right. The cylinder is oriented out of the page. The magnetic field is oriented toward the right, from the north to the south pole, and across the cylinder of fluid. A current-carrying wire runs through the fluid cylinder with current I oriented downward, perpendicular to the cylinder. Negative charges within the fluid have a velocity vector pointing up. Positive charges within the fluid have a velocity vector pointing downward. The force on the fluid is out of the page. An illustration of the right hand rule 1 shows the thumb pointing downward with the current, the fingers pointing to the right with B, and force F oriented out of the page, away from the palm.
Magnetohydrodynamics. The magnetic force on the current passed through this fluid can be used as a nonmechanical pump.

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 [link] .) Existing MHD drives are heavy and inefficient—much development work is needed.

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
Aislinn Reply
cm
tijani
what is titration
John Reply
what is physics
Siyaka Reply
A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
Jude Reply
Can you compute that for me. Ty
Jude
what is the dimension formula of energy?
David Reply
what is viscosity?
David
what is inorganic
emma Reply
what is chemistry
Youesf Reply
what is inorganic
emma
Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes
Adjei
please, I'm a physics student and I need help in physics
Adjanou
chemistry could also be understood like the sexual attraction/repulsion of the male and female elements. the reaction varies depending on the energy differences of each given gender. + masculine -female.
Pedro
A ball is thrown straight up.it passes a 2.0m high window 7.50 m off the ground on it path up and takes 1.30 s to go past the window.what was the ball initial velocity
Krampah Reply
2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
Sahid Reply
you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
Joseph
"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
Ryan
what's motion
Maurice Reply
what are the types of wave
Maurice
answer
Magreth
progressive wave
Magreth
hello friend how are you
Muhammad Reply
fine, how about you?
Mohammed
hi
Mujahid
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
yasuo Reply
Who can show me the full solution in this problem?
Reofrir Reply
Got questions? Join the online conversation and get instant answers!
Jobilize.com Reply

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, College physics (engineering physics 2, tuas). OpenStax CNX. May 08, 2014 Download for free at http://legacy.cnx.org/content/col11649/1.2
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'College physics (engineering physics 2, tuas)' conversation and receive update notifications?

Ask