Magnetic Force on a Current-Carrying Conductor AP Physics C Montwood High School R. Casao.

Magnetic Force on a Current-Carrying Conductor

AP Physics C

Montwood High School

R. Casao

• Just as a force is exerted on a single charged particle when it moves through a magnetic field, a current-carrying wire also experiences a force when placed in a magnetic field.

• A current consists of many charged particles in motion; the resultant force on the wire is the sum of the individual forces on the charged particles constituting the current.

• The magnetic field B is directed out of the page.

• Figure a: no current flows in the wire, so there is no force acting on the wire.

• Figure b: for the current I going from bottom to top, the magnetic force is directed to the right.

• Right-hand rule: fingers of right hand in direction of the velocity of the current (bottom to top); palm faces out away from page in direction of magnetic field B; thumb points to the right.

• Figure c: for the current I going from top to bottom, the magnetic force is directed to the left.

• Right-hand rule: fingers of right hand in direction of the velocity of the current (top to bottom); palm faces out away from page in direction of magnetic field B; thumb points to the left.

• Consider a straight segment of wire of length l and cross-sectional area A carrying a current I in a uniform magnetic field B.

• The magnetic force on a charge q moving with a drift velocity vd is given by: q·v x B.

• The force on the charge carriers is transmitted to the wire through collisions with the atoms making up the wire.

• The total force on the wire is found by multiplying the force on one charge by the number of charges in the segment.

• The volume of the segment is A·l.

• The number of charges in the segment is n·A·l, where n is the number of charges per unit volume.

• Total magnetic force on the wire of length l: Fmag = (q·vd x B)· n·A·l

• Remember that current I = n·q·vd·A. Therefore: Fmag = I·(l x B), where l is a vector in the direction of the current. – The magnitude of l equals the length of the wire

segment.• This equation only works for a straight segment

of wire in a uniform external magnetic field.• The magnetic field produced by the current itself

is also ignored. The wire cannot produce a force on itself.

• Consider a wire of uniform cross-sectional area in an external magnetic field, as shown in the figure.

• Dividing the wire into small segments ds that can are approximately straight, we can describe the magnetic force on a very small segment ds in the presence of a magnetic field B as: BxsdIdF

mag

• dFmag is directed out of the page.

• The magnetic field B can be defined as a measurable force on a current element.– The force is maximum when B is

perpendicular to the current element– The force is zero when B is parallel to the

current element.

• To get the total force F, integrate from one end of the wire (a) to the other end of the wire (b).

b

amagBxsdIdF

• Conclusion: the magnetic force on a curved current-carrying wire in a uniform magnetic field is equal to that on a straight wire connecting the end points and carrying the same current.

• Case I: for a curved wire carrying a current I located in a uniform external magnetic field B.

• The constant magnetic field B can be pulled out in front of the integral:

• The quantity represents the vector sum of all the displacement elements from a to b (the length of the conductor, s).

• Conclusion:

b

amagsdBIdF

b

asd

θsinBsIBxsIF

• Case II: consider a closed loop carrying a current I placed in a uniform external magnetic field B, pull B out in front of the integral.

• The sum of the

displacement vectors

ds is zero for the

closed loop because

the vectors form a

closed loop.

b

amagsdBIdF

• Since , the force Fmag = 0 N.

• Conclusion: the total magnetic force on any closed current loop in a uniform magnetic field is zero.

• Problem Example 29-2: Force on a Semicircular Conductor

• A wire bent into the shape of a semicircle of radius R forms a closed circuit and carries a current I. The circuit lies in the xy plane, and a uniform magnetic field is present along the positive y axis. Find the magnetic forces on the straight portion of the wire and on the curved portion.

0sd

• Straight portion of wire:– Divide the length of the straight portion into equal

elements of length ds.– The angle between each equal element of length ds

and the magnetic field B is 90º.– Using the right hand rule and crossing ds into B

indicates that the direction of the resulting magnetic force is out of the page.

• The net force on the straight portion of the wire is the sum of the force contribution from each element of length ds. Integrating from the left end L to the right end R:

L R

Direction: out of the page. RBI2F

R2BIF

R2s

sBIF

sdBIdF

mag

mag

mag

R

L

R

Lmag

• Semicircle portion:– Divide the semicircle into

small elements of length ds from the right end R to the left end L.

– Each element of length ds is a different angle from the vertical magnetic field B.

– Divide the element of length ds into a component parallel to the magnetic field B and perpendicular to the magnetic field B.

– Remember that the magnetic force is zero for vector components parallel to the magnetic field and maximum for vector components perpendicular to the magnetic field.

– The equation for the perpendicular components is ds·sin .

– The angle will vary from 0 rad to rad for each element of length ds along the semicircular arc.

– arc length equation: s = R·.

• For each element of length ds:

• To get the total force on the semicircular portion, integrate from 0 rad to rad:

dθθsinRBIdF

dθRsdθRs

θsinsdBIdF

mag

mag

RBI211RBIF

1-1RBIF

0cos-πcosRBIF

θcosRBIF

dθθsinRBIF

dθθsinRBIdF

mag

mag

mag

π

0mag

0mag

π

0

π

0 mag

• The direction for the force on the semicircular portion of the wire is into the page.

• Combine the two results. Take out of the page as the positive direction (the positive z-axis) and into the page as the negative direction (the negative z-axis).

This result shows that the net force on a closed loop is 0 N.

N0BRI2BRI2Fmag

• The direction of the magnetic field due to a current carrying element is perpendicular to both the current element ds and the radius vector rhat.

• The right hand rule can be used to determine the direction of the magnetic field around the current carrying conductor: – Thumb of the right hand in the

direction of the current.– Fingers of the right hand curl

around the wire in the direction of the magnetic field at that point.

• The magnetic field lines are concentric circles that surround the wire in a plane perpendicular to the wire.

• The magnitude of B is constant on any circle of radius a.

• The magnitude of the magnetic field B is proportional to the current and decreases as the distance from the wire increases.