SAT II Physics (SN) Episode 2 Part 3 potx

Magnetic Force on Current-Carrying Wires Since an electric current is just a bunch of moving charges, wires carrying current will be subject to a force when in a magnetic field.. The dir

A particle with a positive charge of 3 C moves upward at a speed of 10 m/s It passes

simultaneously through a magnetic field of 0.2 T directed into the page and an electric field

of 2 N/C directed to the right How is the motion of the particle affected?

Answering this question is a matter of calculating the force exerted by the magnetic field and the force exerted by the electric field, and then adding them together The force exerted by the magnetic field is:

Using the right-hand-rule, we find that this force is directed to the left The force exerted

by the electric field is:

This force is directed to the right In sum, we have one force of 6 N pushing the particle to the left and one force of 6 N pushing the particle to the right The net force on the particle

is zero, so it continues toward the top of the page with a constant velocity of 10 m/s

Magnetic Force on Current-Carrying Wires

Since an electric current is just a bunch of moving charges, wires carrying current will be subject to a force when in a magnetic field When dealing with a current in a wire, we

obviously can’t use units of q and v However, qv can equally be expressed in terms of Il, where I is the current in a wire, and l is the length, in meters, of the wire—both qv and Il

are expressed in units of C · m/s So we can reformulate the equation for the magnitude of

a magnetic force in order to apply it to a current-carrying wire:

In this formulation, is the angle the wire makes with the magnetic field We determine the direction of the force by using the right-hand rule between the direction of the current and that of the magnetic field

In the figure above, a magnetic field of T is applied locally to one part of an electric circuit with a 5 resistor and a voltage of 30 V The length of wire to which the magnetic field is applied is 2 m What is the magnetic force acting on that stretch of wire?

We are only interested in the stretch of wire on the right, where the current flows in a downward direction The direction of current is perpendicular to the magnetic field,

which is directed into the page, so we know the magnetic force will have a magnitude of F

= IlB, and will be directed to the right.

We have been told the magnetic field strength and the length of the wire, but we need to calculate the current in the wire We know the circuit has a voltage of 30 V and a

resistance of 5 , so calculating the current is just a matter of applying Ohm’s Law:

Now that we know the current, we can simply plug numbers into the equation for the force of a magnetic field on a current-carrying wire:

The Magnetic Field Due to a Current

So far we have discussed the effect a magnetic field has on a moving charge, but we have not discussed the reverse: the fact that a moving charge, or current, can generate a

magnetic field There’s no time like the present, so let’s get to it

The magnetic field created by a single moving charge is actually quite complicated, and is not covered by SAT II Physics However, the magnetic field created by a long straight wire

carrying a current, I, is relatively simple, and is fair game for SAT II Physics The

magnetic field strength is given by:

The constant is called the permeability of free space, and in a vacuum it has a

For SAT II Physics, it’s not important to memorize this equation exactly It’s more

important to note that the strength of the magnetic field is proportional to the strength of the current and is weaker the farther it is from the wire

The direction of the magnetic field lines are determined by an alternate version of the right-hand rule: if you held the wire with your thumb pointing in the direction of the current, the magnetic field would make a circular path around the wire, in the direction that your fingers curl

EXAMPLE

Two parallel long straight wires carrying a current I stand a distance r apart What force

does one wire exert on the other?

Consider the magnetic field created by the bottom wire as it affects the top wire

According to the right-hand rule, the magnetic field will point out of the page, and will

have a strength of B = ( I)/(2πr).

The force exerted by the bottom wire on the top wire is F = IlB If we substitute in for B

the equation we derived above, we find the force per unit length is:

Using the right-hand rule once more, we find that the force pulls the top wire down toward the bottom wire

We can apply the same equations to find that the top wire pulls the bottom wire up In other words, the two wires generate magnetic fields that pull one another toward each

other Interestingly, the fact that each wire exerts an opposite force on the other is further evidence of Newton’s Third Law.

1 The pointer on a compass is the north pole of a small magnet If a compass were placed next to a bar magnet, as shown above, in what direction would the pointer point?

(A) Out of the page

(B) Into the page

(C) To the left

(D) To the right

(E) In a clockwise pattern parallel to the plane of the page

3 What should one do to maximize the magnitude of the magnetic force acting on a charged particle moving in a magnetic field?

I Maximize the strength of the magnetic field

II Minimize the particle’s velocity

III Ensure that the particle is moving in the same direction as the magnetic field lines (A) I only

(B) I and II only

(C) I and III only

(D) II and III only

(E) I, II, and III

4 What is the magnetic force experienced by a negatively charged particle of 1.0 C that is moving upward at a velocity of 2.0 10 3 m/s in a magnetic field of strength 4.0

10 – 4 T, directed into the page?

(A) 0.8 N to the left

(B) 0.8 N to the right

(C) 2.0 10 – 7 N to the left

(D) 2.0 10 – 7 N to the right

(E) 5.0 10 6 N to the left

5 A charged particle is moving in a circular orbit in a magnetic field If the strength of the magnetic field doubles, how does the radius of the particle’s orbit change?

7 A positively charged particle of 2.0 C moves upward into an area where both a magnetic field and an electric field are acting The magnetic field has a magnitude of 4.0 10 – 4

T and the electric field has a magnitude of 0.1 N/C At what velocity must the particle be moving if it is not deflected when it enters this area?

(A) 4.0 10 – 3 m/s

(B) 125 m/s

(C) 250 m/s

(D) 500 m/s

(E) The particle will be deflected to the left regardless of its velocity

8 A current-carrying wire in a magnetic field is subject to a magnetic force If the current

in the wire is doubled, what happens to the magnetic force acting on the wire?

9 Two wires carry current in opposite directions Which of the following graphs represents the magnetic force acting on each wire?

(A)

(B)

(C)

(D)

(E) There is no net force acting on either wire

10 A current-carrying wire passes through a uniform magnetic field, as shown above At which point is the magnetic field the strongest?

1 B

To solve this problem, it is helpful to remember how the magnetic field lines around a bar magnet look:

The arrows of the magnetic field lines show the direction toward which a north magnetic pole would be attracted Since the compass needle is a south magnetic pole, it’s attracted in the opposite direction of the field lines.

Note that the correct answer is B, and not E The magnet points along the magnetic field lines, and not

straight at the north pole of the magnet.

2 A

This question demands that we apply the right-hand rule backward Force, velocity, and magnetic strength

Let’s imagine the particle at the six o’clock position That means the particle is moving to the left, so stretch your fingers in the leftward direction It’s moving under the influence of a centripetal magnetic force that pulls it in a circle This force is directed toward the center of the circle, so point your thumb upward toward the center of the imaginary clock face To do this, you’ll have to have your palm facing up, and you’ll find that when you curl your fingers around, they point out of the plane of the page That’s the direction of the magnetic field lines.

3 A

proportional to the cross product of v and B, we can maximize F by maximizing v and B, and by ensuring that v and B are perpendicular to one another According to these requirements, only statement I will

maximize the magnetic force: both statements II and III will serve to minimize the magnetic force.

4 B

Since the velocity vector and the magnetic field strength vector are perpendicular, we can calculate the magnitude of the magnetic force quite easily:

The minus sign in the answer signifies the fact that we are dealing with a negatively charged particle That

vector using the right-hand rule: point your fingers upward in the direction of the v vector and curl them downward in the direction of the B vector; your thumb will be pointing to the left Since we’re dealing with a

negatively charged particle, it will experience a force directed to the right.

5 B

If the particle is moving in a circular orbit, its velocity is perpendicular to the magnetic field lines, and so the

magnetic force acting on the particle has a magnitude given by the equation F = qvB Since this force pulls the particle in a circular orbit, we can also describe the force with the formula for centripetal force: F =

mv2/r By equating these two formulas, we can get an expression for orbital radius, r, in terms of magnetic

field strength, B:

Since magnetic field strength is inversely proportional to orbital radius, doubling the magnetic field strength means halving the orbital radius.

6 D

When a charged particle moves in the direction of the magnetic field lines, it experiences no magnetic force,

and so continues in a straight line, as depicted in A and B When a charged particle moves perpendicular to the magnetic field lines, it moves in a circle, as depicted in C When a charged particle has a trajectory that

is neither perfectly parallel nor perfectly perpendicular to the magnetic field lines, it moves in a helix pattern,

as depicted in E However, there are no circumstances in which a particle that remains in a uniform magnetic field goes from a curved trajectory to a straight trajectory, as in D.

If the particle is to move at a constant velocity, then the leftward electric force must be equal in magnitude

to the rightward magnetic force, so that the two cancel each other out:

8 D

The magnetic force, F, due to a magnetic field, B, on a current-carrying wire of current I and length l has a

magnitude F = IlB Since F is directly proportional to I, doubling the current will also double the force.

9 B

Each wire exerts a magnetic force on the other wire Let’s begin by determining what force the lower wire exerts on the upper wire You can determine the direction of the magnetic field of the lower wire by pointing the thumb of your right hand in the direction of the current, and wrapping your fingers into a fist This shows that the magnetic field forms concentric clockwise circles around the wire, so that, at the upper wire, the magnetic field will be coming out of the page Next, we can use the right-hand rule to calculate the direction

of the force on the upper wire Point your fingers in the direction of the current of the upper wire, and then curl them upward in the direction of the magnetic field You will find you thumb pointing up, away from the lower wire: this is the direction of the force on the upper wire.

If you want to be certain, you can repeat this exercise with the lower wire The easiest thing to do, however,

is to note that the currents in the two wires run in opposite directions, so whatever happens to the upper wire, the reverse will happen to the lower wire Since an upward force is exerted on the upper wire,

downward force will be exerted on the lower wire The resulting answer, then, is B

10 C

There are two magnetic fields in this question: the uniform magnetic field and the magnetic field generated

by the current-carrying wire The uniform magnetic field is the same throughout, pointing into the page The magnetic field due to the current-carrying wire forms concentric clockwise circles around the wire, so that

they point out of the page above the wire and into the page below the wire That means that at points A and

B, the upward magnetic field of the current-carrying wire will counteract the downward uniform magnetic

field At points C and D, the downward magnetic field of the current-carrying wire will complement the

downward uniform magnetic field Since the magnetic field due to a current-carrying wire is stronger at

points closer to the wire, the magnetic field will be strongest at point C.

Electromagnetic Induction

CHARGES MOVING IN A MAGNETIC FIELD create an electric field, just as charges

moving in an electric field create a magnetic field This is called electromagnetic induction Induction provides the basis of everyday technology like transformers on

power lines and electric generators

On average, SAT II Physics asks only one question about electromagnetic induction However, less than half of the test takers usually get this question right, so if you get the hang of this material, you’ll be separating yourself from the crowd On the whole, this question will be qualitative, with only a minimum of calculation involved

Motional Emf

Consider the bar in the figure below It has length l and moves at speed v to the right in

magnetic field B, which is directed into the page.

The field exerts a magnetic force on the free electrons in the bar That force is

: using the right-hand rule, you will find that the vector is directed upward along the bar, but since electrons are negatively charged, the magnetic force acting upon them is directed downward As a result, electrons flow to the bottom of the bar, and the bottom becomes negatively charged while the top becomes positively

charged

The separation of charge in the rod creates an electric field within the bar in the

downward direction, since the top of the bar is positively charged and the bottom of the bar is negatively charged The force from the electric field, , pulls negative charges upward while the force from the magnetic field pulls negative charges downward Initially, the magnetic field is much stronger than the electric field, but as more electrons are drawn to the bottom of the bar, the electric field becomes increasingly stronger When the two fields are of equal strength, the forces balance one another out, halting the flow of electrons in the bar This takes place when:

Induced Current and Motional Emf

The electric field in the metal bar causes a potential difference of V = El = vBl If the bar

slides along metal rails, as in the figure below, a closed circuit is set up with current

flowing in the counterclockwise direction, up the bar and then around the metal rail back

to the bottom of the bar This is called an induced current.

The moving bar is a source of an electromotive force, called motional emf, since the

emf is generated by the motion of the bar

The force is defined as:

The magnitude of the induced emf can be increased by increasing the strength of the magnetic field, moving the bar faster, or using a longer bar

EXAMPLE

A bar of length 10 cm slides along metal rails at a speed of 5 m/s in a magnetic field of 0.1

T What is the motional emf induced in the bar and rails?

Now that we’ve defined motional emf, solving this problem is simply a matter of plugging numbers into the appropriate equation:

Faraday’s Law

Moving a conductor through a magnetic field is just one way of inducing an electric current A more common way of inducing current, which we will examine now, is by changing the magnetic flux through a circuit

Magnetic Flux

The magnetic flux, , through an area, A, is the product of the area and the magnetic

field perpendicular to it:

The A vector is perpendicular to the area, with a magnitude equal to the area in question

If we imagine flux graphically, it is a measure of the number and length of flux lines passing through a certain area