Unformatted text preview: ns will be posted at www.prep101.com/solutions 31 ©Prep101 Phys1026 April Exam Solutions Question 8: Solution:
Due to the different directions of current in the loop, one side of the loop is attracted to
the large wire carrying wire, while the other repulsed. Question 9:
Solution:
In this figure, the current in the two wires are in opposite directions. To suspend the top
wire, the magnetic force per unit length must be equal to the wire’s weight per unit
length. Furthermore, we must orient the top wire’s current into the page, in order to get
the force pointing upwards. The magnetic force exerted on a conductor of length l
carrying current It perpendicular to a field of size Bl is F = BlItl. Therefore, the magnetic force per unit length acting on the top wire is: Solutions will be posted at www.prep101.com/solutions 32 ©Prep101 Phys1026 April Exam Solutions Question 10: Solution:
First we find the speed of the proton from its kinetic energy. It is easier to calculate b.)
before a.), therefore we need to know the orbital radius of the proton in the B – field.
B – field directed out of page (y or k axis) As the proton passes through the B – field, the central force is balanced by the magnetic
force, and as the velocity is normal to this field: Therefore, from the path length over which this proton is exposed to the magnetic field
along with the orbital radius, we are able to determine the angle between the initial and
final velocity vectors. It is now straightforward to calculate the xaxis component of the protons’ momentum: 9.11 Answer If the electron is travelling in a circle in a magnetic field, it is undergoing
cyclotron motion. This is where the magnetic field causes it to move in a circle, so the magnetic force is the
centripetal force
FB = FC
qvB = mv2/r
qB = mv/r
v = rqB /m
To find out how often it goes round, we look at Solutions will be posted at www.prep101.com/solutions 33 ©Prep101 Phys1026 April Exam Solutions T = distance / v
= 2π r / v
= 2π r / (rqB/m)
= 2π m / qB
So then the frequency is f = 1/T
f= qB
2πm = 1.6 x 1019 x 1 / (2 x 3.1 x 9.1 x 1031)
= 2.8 x 1010 Hz
The electron completes 2.8 x 1010 orbits every second. The current through a plane is the
flow of charge per second. The electron passes through A 2.8 x 1010 times a second, so
the total current is
I = 2.8 x 1010 x 1.6 x 1019
= 5 x 109 A
The electron passes through B twice an orbit, once in each direction – so the net flow of
charge is zero (just as much flows to the left as to the right). The current through B is
zero.
9.12 Answer e) Magnetic fields only exert forces on moving charges – the sphere is not moving, so there
is no force.
9.13 Answer Strategy develop equation for B at centre of loop
let Bwire = Bloop for a net field strength of zero
use RHR for current direction in the coil Long wire: B = μoIw/2πr
and setting the angle = 2π)
equating the two Loop: B = μoIL/2R ( from the equation for an arc, μoIw/2πr = μoIL/2R
Iw/π(r+R) = IL/R solve for r r = (Iw R) / (π IL) – R Solutions will be posted at www.prep101.com/solutions 3...
View
Full
Document
This document was uploaded on 03/21/2014 for the course PHYSICS 1402 at UWO.
 Winter '14
 Prof.PaulineBarmby
 Physics

Click to edit the document details