Download Report

Homework #8: Magnetic Force and Biot-Savart Law
1. [10 points] Suppose that an electron is released from rest in a region where there is a
uniform electric field in the z-direction and a magnetic field in the x-direction. Determine
the trajectory of the electron if it starts at the origin with velocity.
)̂
a. ⃗( ) (
)̂
b. ⃗( ) (
)( ̂
c. ⃗( ) (
̂)
Answer: Since the electron is in the presence of an electric field and a magnetic field, we apply
Newton’s 2nd law to the Lorentz force law
⃗
( ⃗⃗
⃗
⃗
⃗⃗ )
Since there is no force in the x-direction, then the velocity can be written as
⃗
(
)
Writing Newton’s 2nd law in component form gives us
( )
(
where
)
. Differentiating (1) with respect to t and substituting (2) gives
(
Letting
( )
)
gives
( )
To solve this, we can use the method of undetermined coefficients. The complementary solution
is given by solving the corresponding homogeneous ODE to (3), which gives
The particular solution should have the form
, where
is the undetermined
coefficient. Substituting the particular solution into the ordinary differential equation gives
. Therefore, the general solution to this differential equation is given by
( )
( )
( )
Therefore, the z solution is given by
( )
Therefore, the general solution to (1) and (2) is given by
( )
( )
( )
In order to determine
, we need to know the initial conditions. Since the electron
starts from rest, this implies that ( )
( ). Note that ( )
implies that
.
Note that ( )
implies that
. To specify the coefficients further, we need to use
the information regarding the initial velocity vector.
(a) Note that
( )
( )
implies that
implies that
. Note that
. Therefore, for this initial velocity profile,
( )
( )
This indicates that the particle moves in a straight line at constant speed.
(b) Note that ( )
implies that
. Since
, this implies that
. Note that ( )
implies that
. Therefore, for this initial velocity profile,
( )
( )
[
]
Rearranging the above equations gives us
( )
( )
Squaring both equations and adding them together gives
[ ( )
This is a circle of radius
]
[ ( )
]
whose center is (
positive y-direction at constant speed, as shown below.
(
)
)
(
), moving in the
( )
(c) Note that that
that ( )
implies that
Note that
. Therefore, we have
implies that
( )
[
]
( )
Rearranging the above equations gives us
( )
( )
Squaring both equations and adding them together gives
[ ( )
This is a circle of radius
positive y-direction.
]
( )
whose center is (
(
)
)
(
), moving in the
2. Consider a rectangular loop of wire, supporting a mass m, which hangs vertically with
one end in a uniform magnetic field ⃗⃗, which points into the page.
a. [3 points] What current I in the loop would exactly balance the weight of the
mass?
b. [7 points] Suppose we now increase the current such that the magnetic force
exceeds the weight of the mass. This causes the loop to rise and thus, lifts the
mass. What is the work done to lift the wire? What force is doing work here?
Defend your answer.
Answer: A schematic diagram of this scenario is shown below
(a) The magnetic force on a current carrying wire is given by ⃗
where a is the width
of the loop. In order to balance the weight of the object, we must have
(b) As the loop starts to rise, the charges in the wire are no longer moving horizontally – their
velocity now acquires an upward component u, the speed of the loop, in addition to the
horizontal component w associated with the current as shown below.
The magnetic force, which is also perpendicular to the velocity, no longer points up, but
tilts back according to the right hand rule. It is perpendicular to the net displacement of
the charge (which is in the direction of v), and thus, does no work on the charge. The net
vertical force on all charge in the upper segment of the loop is
,
but now it also has a horizontal component (
), which opposes the flow of the current.
Thus, whatever initiates the current (i.e. a voltage source like a battery) must do work
against the backward component of the magnetic force. The total horizontal force on the
top segment is given by
. In a time dt, the charges move a horizontal
distance
, so the work done by the voltage source is given by
∫
3. Consider a circular wire loop of radius R located in the yz plane and carrying a steady
current I, as shown below.
a. [7 points] Calculate the magnetic field at an axial point P a distance x from the
center of the loop.
b. [3 points] Show that the magnitude of the magnetic field far from the loop is
given by
(
Here,
) is defined as the magnetic moment of the loop.
Answer:
By symmetry, the current in any element on one side of the loop sets up a perpendicular
component of the magnetic field that cancels the perpendicular component set up by the current
through the elements opposite of it. Therefore, the resultant field at point P must be along the x
axis. Moreover, note the every length segment ⃗ is perpendicular to the separation vector at the
location of the element. Therefore, we have
⃗⃗
⃗
̂
̂
̂
̂
√
(
̂
)
Therefore, the total magnetic field is given by
⃗⃗
∮
(
)
(
For points far away from the loop (i.e.
̂
̂
(
)
̂
), we have
)
∮
̂
(
)
(
)
⃗⃗
(
)
̂
̂
(
(
̂
)
̂
̂
)
4. Magnetic Force Exerted on Current Carrying Loops
a. [5 points] Find the force on a square loop near an infinite straight wire. Assume
that both the loop and the wire carry a steady current I.
b. [5 points] Find the force on the triangular loop. Assume that the loop and wire
carry a steady current I.
Answer: The magnetic field associated with an infinite wire carrying a steady current I is given
by
where s is the vertical distance from the wire. We can use this result to determine the force on
each square loop.
(a) Since the sides of the current loop are both a distance s from the wire, the magnitude of
the magnetic force on the sides is equal. However, since the current moves in opposite
directions, the magnetic force on each side is equal and opposite. Therefore, the forces on
the two sides cancel. On the bottom of the loop, we have
(
)
At the top of the loop, we have
(
)
(
(
)
)
(
)
Since the force on the bottom of the loop points upward and the force on the top of the
loop points downward by the right hand rule, we have
(
)
(
)
(b) Consider the schematic below
The force on the bottom is the same as before,
On the left and right of the loop, we have ⃗⃗
the left side of the loop is given by
⃗
( ⃗
⃗⃗ )
[
̂
̂ . The force on a line element on
̂]
̂
[
̂]
(
̂
̂)
(
̂
̂)
The force on a line element on the right side of the loop is given by
⃗
( ⃗
⃗⃗ )
[
̂
̂]
̂
[
̂]
Thus, the x-component cancels the corresponding term and the net force will be pointing
only in the y-direction. Therefore, the total force due to both sides of the current loop will
be
(
)∫
√
(
)∫
√
√
√
(
√
√
)
(
The net force on the triangular loop is given by
√
(
√
)
[
√
(
√
)]
√
)
5. [10 points] A semicircular wire carries a steady current I as shown below. Show that the
magnetic field at a point P on the other semicircle is
[
(
)
( )
]
Answer: A schematic of this configuration is given below
Let the source point and field point be
⃗
⃗
̂
̂
⃗
̂
̂
(
̂) To determine the magnetic field
using the Biot-Savart law, we first need to determine ⃗ ̂ where ̂ is the separation
vector between the field point and the source point. By definition of the cross product, we
have
Let the line element be
̂
⃗
̂
̂
̂
̂
|
|
(
)
[
[
(
)
(
)
(
] ̂
)] ̂
[
(
)] ̂
Using the Biot-Savart law (in terms of the separation vector), we now have
⃗ ̂
[
(
)]
⃗⃗
∫
̂∫
[
(
)]
(
(
)
)
̂∫
̂∫
[
√
(
(
)]
)
[
(
)
( )
] ̂
Bonus [7 points]: Using the Biot-Savart law, show that the magnetic field at a point
axis of the rotating sphere, as shown below, is given by
⃗⃗
Here,
on the
̂
is the angular speed of a charge element around the z-axis
Hint: The following integrals may be helpful
∫ √
[(
∫
∫
)
) ]
(
√
(
√
)
Answer: We can think of a sphere as a large collection of rings. Therefore, if we find the
magnetic field due to a ring due to a point on the z-axis, we can generalize this result for the
(
) . The time for one revolution is
sphere. The total charge on the shaded ring is
so the current in the ring is
. The magnetic field of this ring for
points on the axis is given by
∫
(
)
[
(
√
For a sphere, we can slice the sphere into slabs of thickness
result where
,
, and
)
]
and we can use the above
. Therefore, we have
∫
If we let
(
[√
)
(
√
)]
, then we have
∫ [√
(
)
(
)
[∫ √
(
)∫
]
√
(
)∫
∫
√
√
∫
]
To solve this integral expression, we can use the following integrals
∫ √
[(
∫
∫
[
[(
[
Since
(
(
(
(
) ]
(
√
√
)
)
) ]
)
(
)(
)
(
), then we have
⃗⃗
̂
)
(
)[
)
(
]
)]
]