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Physics 121 Practice Problem Solutions 10 Magnetic Fields from Currents (Biot-Savart and Ampere

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Title: Physics 121 Practice Problem Solutions 10 Magnetic Fields from Currents (Biot-Savart and Ampere


1
Physics 121 Practice Problem Solutions 10
Magnetic Fields from Currents(Biot-Savart and
Amperes 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
  • Amperes Law
  • Solenoids and Toroids

2
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 mT.
3
(No Transcript)
4
PROBLEM 121P10-8P Use the BiotSavart 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.
5
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?
6
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.)
7
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
8
PROBLEM 121P10-39P The figure shows a cross
section of an infinite conducting sheet carrying
a current per unit x-length of l the current
emerges perpendicularly out of the page. (a) Use
the BiotSavart 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 m0l 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
sheet
Length L
The enclosed current ienclosed l 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 m0ienclosed
m0 l L implies B m0 l /2
9
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.
10
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.
  • Same result as for long solenoid with length
    taken to be circumfrence, i.e.,
  • b) Find B field outside
  • Circular Amperian loop as before, but outside the
    toroid.
  • Now net current enclosed 0, so Amperes Law
    argument yields B 0
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