magnetic fields and currents - smu physics · a wire carrying a current i=6.0a is partly bent into...

Physics 1308: General Physics II - Professor Jodi Cooley

Sir Joseph John Thomson 18 December 1856 – 30 August 1940

Magnetic Fields and Currents

Welcome Back to Physics 1308


Announcements

• Assignments for Thursday, November 1st:

- Reading: Chapter 30.1 - 30.5

- Watch Videos:

- https://youtu.be/y56A0BZGuuU — Lecture 19 -Magnetic Induction (00:00 - 56:54)

• Homework 10 Assigned - due before class on Tuesday, November 6th.

https://youtu.be/y56A0BZGuuU


Review Question 1A rectangular loop is placed in a uniform magnetic filed with the plane of the loop perpendicular to the direction of the field. If a current is made to flow through the loop in the sense shown by the arrows, the field exerts on the loop:

A)a net force. B) a net torque C) a net force and a net torque D)neither a net force nor a net torque.

According to the RHR, all for sides of the loop are subject to forces that are directed outward, with forces on each side canceling that on the opposite side. The magnetic moment is parallel to the magnetic flux. Hence, the torque is zero.


Key Concepts• Electric charge exists in two kinds, positive and negative. These exert influence on

each other via a field of force.

• The electric field is a conservative force field, with associated concepts of potential energy and the new concept of “electric potential”. Fields accelerate charges, and this is the basis of circuits. There are relationships between fields and the energy stored and release in circuits (e.g. in capacitors and resistors).

• Magnetic fields, fields of force exerted by certain materials such as magnetite and iron, also affect electric charges. However, they do so at right angles to the field lines and the direction of motion of the charges. Magnetic fields are caused by the motion of electric charge.


Key ConceptsBiot-Savart Law:

ȓ is a unit vector that points from ds towards point P


Key ConceptsSolenoids:

• The solenoid’s magnetic field is the vector sum of the fields produced by the individual turns (windings) that make up the solenoid.

• For points very close to a turn, the wire behaves magnetically almost like a long straight wire, and the lines of B there are almost concentric circles.

• Field tends to cancel between adjacent turns. • At points inside the solenoid and reasonably far from the wire, B

is approximately parallel to the (central) solenoid axis. • Inside a long solenoid carrying current i, at points not near its

ends, the magnitude B of the magnetic field is


Question 1The drawing shows a rectangular wire loop that has one side passing through the center of a solenoid. Which one of the following statements describes the force, if any, that acts on the rectangular loop when a current is passing through the solenoid.

A) The magnetic force causes the loop to move upward. B) The magnetic force causes the loop to move

downward. C) The magnetic force causes the loop to move to the

right. D) The magnetic force causes the loop to move to the

left. E) The loop is not affected by the current passing

through the solenoid or the magnetic field resulting from it.


Instructor Problem: Line of Current Revised

Consider a vertical segment of wire carrying a current, i, from the top to the bottom of the screen. The segment is of length L. What is the magnetic field at a point P, a distance D to the right of the line, if the point is halfway along the length of the wire?


In this problem, we need to use the Biot-Savart Law to relate current to the resulting magnetic field a distance D away.

d ~B =µ0

4⇡

Id~L⇥ r

r2

Similar to Coulomb’s Law — we need to assess the following:

1. Choose appropriate coordinate systems. Straight lines bene t best from Cartesian Coordinates, while circles typically are solved best via Polar Coordinates.

2. Write dL, the length component of the current element, in terms of coordinates, constants, and differentials.

3. Write the vector r in terms of coordinates and constants. Use, as always, the convention that this vector points FROM the source of the field TO the place where we are measuring/calculating/experiencing the field.

4. Neet to get from r.r


Since this is a straight line, let’s use cartesian coordinates, with the x-axis along the charge line and the origin at point P.

d~L = dy(�j)

First let’s find an expression for the line element dL.

Now let’s move onto r.

~r = Di� yj

r2 = D2 + y2

r =(Di� yj)pD2 + y2

Recall:

r =~r

r

i r


Now we have all the pieces to assemble the Biot-Savart Law.

d ~B =µ0

4⇡

Id~L⇥ r

r2

=µ0

4⇡

Idy(�j ⇥ r)

D2 + y2

=µ0

4⇡

Idy

(D2 + y2)3/2(�j ⇥ (Di� yj))

Let’s assess the cross products in the parentheses:

j ⇥ i

j ⇥ j

= �k

= 0

=µ0

4⇡

Idy

(D2 + y2)

✓� j⇥

✓(Di� yj)pD2 + y2

◆◆

r =(Di� yj)pD2 + y2

d~L = dy(�j)


So, this simplifies our problem.

d ~B =µ0

4⇡

IDdy

(D2 + y2)3/2k

The final step is to integrate this from the bottom of the line to the top. Since, we chose zero point is in the middle of the line, our we integrate from -L/2 to +L/2.

~B =

Z +L/2

�L/2

µ0

4⇡

IDdy

(D2 + y2)3/2k

=µ0ID

4⇡(k)

Z +L/2

�L/2

dy

(D2 + y2)3/2

~B =µ0ID

4⇡(k)

1

D2

✓L/2p

D2 + (L/2)2� (�L/2)p

D2 + (�L/2)2

◆

~B =µ0I

4⇡D

LpD2 + (L/2)2

(k)

=µ0ID

4⇡(k)

✓y

D2p

D2 + y2

◆��L/2

L/2-L/2


Student Problem: Arc of Current Carrying Wire

A wire carrying a current I=6.0A is partly bent into a semi-circle, as shown above. The semi-circle has a radius of R = 0.050m, and sits in between two very long straight sections of wire. The current flow is continuous from left to right. What is the strength of the magnetic field, B, at the point, P, at the center of the semi-circle? HINT: use polar coordinates!


First, recall that a magnetic field is a vector. If we have three magnetic fields, all acting on a point P in space, then the total magnetic field is given by

~Btotal =3X

i=1

~Bi

The Biot-Savart law lets us determine the magnetic field due to complex, current carrying shapes by considering the shape to be made of finite elements, each generating a piece of the magnetic field.

d ~B =µ0

4⇡

Id~L⇥ r

r2

To obtain the total magnetic field, we integrate (or sum up) the elements.

~Btotal =3X

i=1

~Bi.

d ~B =µ0

4⇡

i d~L⇥ r

r2,

µ0 µ0 = 4⇡⇥10�7T ·A/m i ~LL ~r

~B =

ˆd ~B =

ˆ max

min

µ0

4⇡

i d~L⇥ r

r2.

i

i d~Li


This problem can be broken into 3 pieces. The current, i, (1) flowing from left to right on a straight line of conductor, then (2) clockwise around a semi-circle of conductor, and finally (3) again straight from left to right through a third straight piece of conductor.

First we pick a coordinate system. Since P lies in the same line as both pieces, the cartesian coordinate system is convenient.

Let’s start with piece (1) and (3) since they are very similar to each other.

Now choose idL. This points in the direction of flow which is to the right (positive x direction).

dL

r

Now lets draw the vector r. What do we notice?It is parallel to the idL vector. Hence, via the cross product in Biot-Savart, it will contribute zero. B1 = B3 = 0 T.


Now consider piece 2.

dL

Since this is a semi-circle, it is best to use circular coordinates (polar). We can convert back to cartesian later using the relationships x = Rcosθ and y = Rsinθ.

Let’s start by looking at r. Since we are using polar coordinates we can already write

r2 = R2

What about our unit vector?r = cos ✓i+ sin ✓j

But — we don’t really need this because we already know that dL and make a 90 degree angle with each other

r


So we can write:

|id~L⇥ r| = idL sin 90� = idL

Using the right-hand-rule, we find the vector points into the page.

The final piece is dL which is equal to an arc length ds.

d ~B =µ0

4⇡

id~L⇥ r

r2

=µ0

4⇡

ids

R2(�k)


Now integrate:

~B2 =

Z 12 (2⇡R)

0

µ0

4⇡

ids

R2(�k)

=µ0

4⇡

i

R2(�k)

Z ⇡R

0ds

=µ0

4⇡

i

R2(�k)

✓s

��⇡R

0

◆

=µ0

4⇡

i

R2(�k)(⇡R)

Since this is the ONLY magnetic field contributing to the total flux… we have

~B =µ0i

4Rk �! ~B = 3.8⇥ 10�5 T


The End for Today!

magnetic fields and currents - smu physics · a wire carrying a current i=6.0a is partly bent into...

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