Physics 1308: General Physics II - Professor Jodi Cooley

Sir Joseph John Thomson 18 December 1856 – 30 August 1940

Magnetic Fields

Welcome Back to Physics 1308

Physics 1308: General Physics II - Professor Jodi Cooley

Announcements

• Assignments for Tuesday, October 30th:

- Reading: Chapter 29.1 - 29.3

- Watch Videos:

- https://youtu.be/5Dyfr9QQOkE — Lecture 17 - The Biot-Savart Law

- https://youtu.be/0hDdcXrrn94 — Lecture 17 - The Solenoid

• Homework 9 Assigned - due before class on Tuesday, October 30th.

Physics 1308: General Physics II - Professor Jodi Cooley

Physics 1308: General Physics II - Professor Jodi Cooley

Review Question 1Consider the two rectangular areas shown with a point P located at the midpoint between the two areas. The rectangular area on the left contains a bar magnet with the south pole near point P. The rectangle on the right is initially empty. How will the magnetic field at P change, if at all, when a second bar magnet is placed on the right rectangle with its north pole near point P?

A) The direction of the magnetic field will not change, but its magnitude will decrease.

B) The direction of the magnetic field will not change, but its magnitude will increase.

C) The magnetic field at P will be zero tesla. D) The direction of the magnetic field will change and its magnitude will increase. E) The direction of the magnetic field will change and its magnitude will decrease.

Physics 1308: General Physics II - Professor Jodi Cooley

Review Question 2An electron traveling due east in a region that contains only a magnetic field experiences a vertically downward force, toward the surface of the earth. What is the direction of the magnetic field?

A)upward, away from the earth B) downward, toward the earth C) due north D)due west E) due south

Note: Don’t forget to flip your thumb at the end due to the fact that the electron has a negative charge!

Physics 1308: General Physics II - Professor Jodi Cooley

Key ConceptsTorque on a Current Loop:

Various magnetic forces act on sections of a current carrying coil lying in a uniform external magnetic field, but the net force is zero.

The net torque acting on the coil has a magnitude given by

⌧ = NiAB sin ✓

where N is the number of turns in the coil, A is the area of each turn, i is the current i, the magnetic field magnitude B, and θ is the angle between the magnetic field B and the normal vector to the coil n.

Physics 1308: General Physics II - Professor Jodi Cooley

Key ConceptsMagnetic Dipole Moment:A coil (of area A and N turns, carrying current i) in a uniform magnetic field B will experience a torque τ given by

where µ is the magnetic dipole moment of the coil, with magnitude µ = NiA and direction given by the right-hand rule.

The orientation energy of a magnetic dipole in a magnetic field is

If an external agent rotates a magnetic dipole from an initial orientation θi to some other orientation θf and the dipole is stationary both initially and finally, the work Wa done on the dipole by the agent is

Physics 1308: General Physics II - Professor Jodi Cooley

Instructor Problem: The Magnetic Dipole Moment of a Coil

A rectangular coil of dimensions 5.40 cm x 8.50 cm consists of 25 turns of wire and carries a current of 15.0 mA. A 0.350 T magnetic field is applied parallel to the plate of the coil.

A) Calculate the magnitude of the magnetic dipole moment of the coil. B) What is the magnitude of the torque acting on the loop?

Physics 1308: General Physics II - Professor Jodi Cooley

Part A:

Given: ` = 8.50 cm

w = 5.40 cm

N = 25

B = 0.350 T

The magnetic moment of a coil is given by

µ = NIA

i = 15 mA

= (25)(15⇥ 10�3 A)(0.054 m⇥ 0.85 m)

µ = 1.72⇥ 10�3 A ·m2

Part B:

The magnitude of the torque is given by

|~⌧ | = µB sin ✓ = (1.72⇥ 10�3 A ·m2)(0.350 T ) sin (90)

|~⌧ | = 6.02⇥ 10�4 N ·m

~⌧ = ~µ⇥ ~B �!

Physics 1308: General Physics II - Professor Jodi Cooley

Student Problem: Rotating a CoilConsider the loop of wire in the Figure shown. Imagine it is pivoted along side 4, which is parallel to the z-axis and fastened so that side 4 remains fixed and he rest of the loop hangs vertically in the gravitational field of Earth but can rotate around side 4. The mass of the loop is 50.0 g, and the sides are of lengths a = 0.200 m and b = 0.100 m. The loop carries a current of 3.50 A and is immersed in a vertical uniform magnetic field of magnitude 0.0100 T in the positive y-direction. What angle does the plane of the loop make with the vertical?

Physics 1308: General Physics II - Professor Jodi Cooley

Given: m = 50.0 g

a = 0.200 m

b = 0.100 m

~B = 0.0100 T (j)

i = 3.50 A (CCW )

We are told the loop hangs vertically so that it can pivot around side 4. Let’s take an edge view of the loop — sketch of this.

The direction of the magnetic moment is normal to the plane of the coil.

Physics 1308: General Physics II - Professor Jodi Cooley

The magnetic torque is given by the equation

~⌧B = ~µ⇥ ~B

Employ our cross product rules. Recall, only components that are perpendicular to the magnetic field contribute. So, we need the x-component of the magnetic moment. In this calculation we will be crossing i and j.

θ

|~⌧B | = �µB sin (90� � ✓)(k) = �iAB cos ✓ (k)

The gravitational torque can be is given by the equation

~⌧g = ~r ⇥m~g = mgb

2sin ✓ k

Since the system is in equilibrium, the torques must be equal to zero.X

⌧ = 0 = �iAB cos ✓ k +mgb

2sin ✓ k

iAB cos ✓ = mgb

2sin ✓ �! tan ✓ =

2iAB

mg

So the resultant vector will be in the k direction.

✓ = tan�1 (2)(3.50 A)(0.200 m)(0.100 )

(0.050 kg)(9.80 m/s2)

✓ = 16.4�

Physics 1308: General Physics II - Professor Jodi Cooley

The End for Today!