Chapter 20
The Giancoli onLine Tutor

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Magnetism

The material in this Chapter is best presented at a higher mathematical level where indeed solving pronlems can be complex.At this level, however, the problems are, of necessity, pretty simple consisting mostly of simply plugging into the appropriate equations. The strategy, then, is to be able to identify the equation that is needed in each case .

I. Wires producing magnetic fields

A long straight wire

B=(m_o/2p)I/r
where m_o is the permeability of free space and equals 4px10^-7 so (m_o/2p)= 2x10^-7. B is the magnetic field surrounding the wire, I is the current carried in the wire and r is the perpendicular distance from the wire to the point where B is to be evaluated. The field is counter-clockwise if the current is 'towards you' and clockwise otherwise.

Interactive Example
What is the magnetic field 20 cm from a wire carryinc a current of 50 A? Enter your answer below and click outside.

The solenoid

A solenoid is wire wrapped around a cylinder. The figure shows the longitudinal cross-section (a cut through its length). The

represents current coming towards you as it winds around the cylinder and

is the current as it goes away. As the equation shows the field is constant (technically, 'uniform') through out the interior of the solenoid depending on n,( the total number of turns, N, divided by the length, L, of the solenoid) and I the current through the wires.

A variation on this is to fill the interior of the cylinder with something usually 'soft' iron, creating an electromagnet. The new field in the interior and at the ends is now given by

B= mnI
where m is the magnetic permeability of the interior material and has a complicated behavior. You should not encounter this for the most part at this level.

Interactive Example
Using 3.14 for p what is the magnetic field inside a 20 cm long a solenoid with 800 wraps and a current of 2A? Enter your answer below and click outside.

II. Magnetic fields produce a force on wires carrying electric current

The direction of the force is always perpendicular to I and B and its 'sense' (in or out) is determined by the right-hand-rule as discribed in the text. As drawn with the magnetic field from left to right the direction of the force is into the page if sinq is positive (as measured from East, that is, right) and out when sinq is negative (third and forth quadrants).

Special Cases:

Forces between 2 parallel wires
This is essentualy a combination of the upper two points. Wires carrying current both produce and respond to magnetic fields, so parallel wires (with currents) exert a force on each other. Combining the equations we obtain:

F/l=(m_o/2p)I₁I₂/L

where F/l is the force per unit length on each wire . I₁ and I₂ are the mutual currents and L the separation. The force is attactive if the currents are in the same direction and repulsive otherwise.

Interactive Example
Two parallel 2-meter long wires carrying identical currents I are separated by 2 mm and experience a force of .0128 N. What is the current? Enter your answer below and click outside.
Torque on a coil of wire
A coil of wire has a magnetic dipole moment, M, its magnitude is equal to the product of a physical factor, (the numer turns of wire in the coil,N) an electrical factor ( the current, I) and a geometrical factor (the area enclosed by the coil, A). Its direction is given by the right-hand-rule as discribed in the text. The torque, t, on the coil is given by

t= MBsinq

where q is the angle betweem the magnetic dipole moment vector and the magnetic field.

A coil of current carrying wires is used in many devices as described in the text. Most problems amount to adjusting one of the factors in (M=) NIA so that M remains constant when another factor changes.

Interactive Example
What is the maximun torque on a 10 coil current loop with a radius of 10 cm carrying 6 A in a 1 T magnetic field?
Enter you answer here and click the clover to try your luck:

III. Magnetic fields produce a force on moving charges

The force is perpendicular to both v and B with a sense given exactly as with a current carrying wire above if the charge is positive. A negative charge, of course, reverses the direction of the force. There are three classes of problems:

Motion in a constant magnetic field.

If the velocity of the charge, q, and mass,m, is perpendicular to the magnetic field the motion will be in a circle with a radius given by

r= mv/qB.

The time it takes to make one revolution is
T= 2pm/qB

If the velocity is not perpendicular to B replace v in the above with the component of v perpendicular to B, that is, vsinq. The parallel component of v, v_||= vcosq is unaffected and the charge undergoes a spiral with each turn of the spiral separated by v_||T and the radius as given above.

Interactive Example
An electron travels at 4x10⁷ m/s perpendicular to a .02 T magnetic field. What is the radius of its circular path to the nearest cm? Enter your answer below and click outside.
Enter you answer here and click the clover to try your luck:

Magnetic force balancing another force.

The magnetic force can be used to balence another force, such as, the weight (mg) or the electric force (qE). In either case just set qvBsinq equal to the force. (Of course, the directions of the velocity, magnetic field and the other force must be properly oriented so that the two forces oppose each other.

Interactive Example
How fast must a particle with charge to mass (q/m) ratio of 1x10^-4, move to the nearest m/s, perpendicularly to a magnetic field of 0.05 T for the magnetic force on it to equal its weight? (Use 10 for g.) Enter your answer below and click outside.
Enter you answer here and click the clover to try your luck:

A combination of the two.

The classic example is the Mass Spectrograph.

The

part of the apparatus is a 'velocity selector' . With a magnetic field in the green region into the screen a positive charge entering the green region from horizontally from the left will expeirence a magnetic forceup and an electric force down. Only those for which the two forces exactly balence will be undeflected and exit through the hole on the right. The condition for this is v=E/B.

The blue region contains another magnetic field, B', directed out of the screen which bends the particle in a circle with the radius given above. Using this with the known velocity the mass can be determined. The result being

m=qBB'r/E

Interactive Examples
A charged particle travelling at 2x10⁶ m/s passes undeflected through a velocity selector that has an electric field equal to 150000 V/m. What is the strength of the magnetic field ? Enter your answer below and click outside.

The particle (m= 2x10^-26, q=1.6x10^-19) now enters the blue region which has a magnetic field of .075 T. To the nearest hundredth of a meter what is the radius of its circular path ? Enter your answer below and click outside.