Fall 2012

Physics 121 Practice Problem Solutions 10 Magnetic Fields from Currents

(Biot-Savart and Ampere’s Law)

Contents:

121P10 - 1P, 5P, 8P, 10P, 19P, 29P, 39P, 50P

• Introduction

• Magnetic Field is due to a Currents

• Biot-Savart Law

• A Loops of Current

• Magnetic Dipole Moment

• Field due to a long Wire

• Force Between Two Parallel Wires Carrying Currents

• Ampere’s Law

• Solenoids and Toroids

Fall 2012

PROBLEM 121P10-1P: A surveyor is using a magnetic compass 6.1 m below a power line in which there is a steady current of 100 A. (a) What is the magnetic field at the site of the compass due to the power line? (b) Will this interfere seriously with the compass reading? The horizontal

component of Earth's magnetic field at the site is 20 µµµµT.

Fall 2012

PROBLEM 121P10-5P*: A particle with positive charge q is a distance d from a long straight wire that carries a current i; the particle is traveling with speed v perpendicular to the wire. What are the direction and magnitude of the force on the particle if it is moving (a) toward and (b) away from the wire?

Fall 2012

PROBLEM 121P10-8P*: Use the Biot–Savart law to calculate the magnetic field B at C, the common center of the semicircular arcs AD and HJ in the figure . The two arcs, of radii R2 and R1, respectively, form part of the circuit ADJHA carrying current i.

Fall 2012

PROBLEM 121P10-10P: The wire shown in the figure carries current i. What magnetic field B is produced at the center C of the semicircle by (a) each straight segment of length L, (b) the semicircular segment of radius R, and (c) the entire wire?

Fall 2012

PROBLEM 121P10-19P: The figure shows a cross section of a long thin ribbon of width w that is carrying a uniformly distributed total current i into the page. Calculate the magnitude and direction of the

magnetic field B at a point P in the plane of the ribbon at a distance d from its edge. (Hint: Imagine the

ribbon to be constructed from many long, thin, parallel wires.)

Fall 2012

PROBLEM 121P10-29P*: In the figure, the long straight wire carries a current of 30 A and the rectangular loop carries a current of 20 A. Calculate the resultant force acting on the loop. Assume that a = 1.0 cm, b = 8.0 cm, and L = 30 cm.

UP

Fall 2012

PROBLEM 121P10-39P: The figure shows a cross section of an infinite conducting sheet carrying a

current per unit x-length of λ; the current emerges perpendicularly out of the page. (a) Use the Biot–Savart law and symmetry to show that for all points P above the sheet, and all points P´ below it, the magnetic field B is parallel to the sheet and directed as shown. (b) Use Ampere's law to prove that B =

½ µ0λ at all points P and P´.

The infinite sheet means the field above and below are parallel to the plane of the sheet as shown in the sketch. Choose a rectangular Amperian loop as shown below, centered on the sheet

∫∫∫∫ Amperes law: B.ds = µ0ienclosed

for a closed “Amperian” loop

sheet

The enclosed current ienclosed = λ L. The field has the same magnitude on upper and lower horizontal segments and equals + BL for each. For the vertical segments B.ds = 0, so there is no contribution.Collecting:

2BL = µ0ienclosed = µ0 λ L implies B = µ0 λ /2

Length L

Fall 2012

PROBLEM 121P10-50: The figure shows an arrangement known as a Helmholtz coil. It consists of two circular coaxial coils, each of N turns and radius R, separated by a distance R. The two coils carry equal currents i in the same direction. Find the magnitude of the net magnetic field at P, midway between the coils.

Fall 2012

PROBLEM 121P10-Torr The figure shows a cross section of a toroid – a solenoid consisting of N turns of wire bent around into a circle of radius r. A symmetry argument has already been used to show that the field lines inside the toroid are circles.

a) Use Ampere's law to find the magnitude of the field inside the toroid and b) Find the magnitude of the field outside.

a) Find the magnitude of B field inside

� Draw an Amperian loop parallel to the field, with radius r (inside the toroid)

� The toroid has a total of N turns

� The Amperian loop encloses current Ni.

� B is constant on the Amperian path.

iNirBsdB enc 002 µµµµ====µµµµ====ππππ====⋅⋅⋅⋅∫∫∫∫rr

toroid inside r

iNB

ππππ

µµµµ====

2

0

� Same result as for long solenoid with length taken to be circumfrence, i.e.,

0 length)t (turns/uni in B r

N n µµµµ====

ππππ≡≡≡≡ ⇒⇒⇒⇒

2

b) Find B field outside

• Circular Amperian loop as before, but outside the toroid.

• Now net current enclosed = 0, so Ampere’s Law argument yields B = 0 outside 0 B ====