(Based (1with the familiar notation), on the infinitely small magnetic force dF, exerted on an infinitely small charge dq, moving with the velocity v, i.e. dF v dq x B, show that the magnitude dF(x), of the force exerted on the infinitely small piece of payload of width |dx], around the location |x, can be expressed as dF(x)= idx| B(x).
(Based (1with the familiar notation), on the infinitely small magnetic force dF, exerted on an infinitely small charge dq, moving with the velocity v, i.e. dF v dq x B, show that the magnitude dF(x), of the force exerted on the infinitely small piece of payload of width |dx], around the location |x, can be expressed as dF(x)= idx| B(x).
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Question
![c).
Show thus that the magnetic force exerted on the unit length of payload, can be
written as
Họ iD
|dx| 1(D? -x?)
dF(x)
(2)
d,. As you can see right away, B(x) turns out to be infinite at x = -D and x = D. How
coie that a finite current can produce an infinite magnetic force? (What is inadmissible in
the derivation of the above expression?)
We overlook the foregoing problem, and we propose to calculate dF(x)/dx], at x=0.
What should be i, if we wish to obtain a force per unit length, just to overcome Earth's
acceleration, g=10 m/s?, for a payload of mass m=200 kg, given that D = 1 m? Hint: Find
the weight per unit length (mg/2D) and equate it to the right hand side of Eq.(2). You will
see that i, is huge, and this is why this technique is not appropriate to launch anything to
space, from Earth.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F35f75a04-2374-43e5-9e66-5970c48e4b4c%2F860cf77b-f427-4b61-946f-f002d656a594%2F7bj18n_processed.jpeg&w=3840&q=75)
Transcribed Image Text:c).
Show thus that the magnetic force exerted on the unit length of payload, can be
written as
Họ iD
|dx| 1(D? -x?)
dF(x)
(2)
d,. As you can see right away, B(x) turns out to be infinite at x = -D and x = D. How
coie that a finite current can produce an infinite magnetic force? (What is inadmissible in
the derivation of the above expression?)
We overlook the foregoing problem, and we propose to calculate dF(x)/dx], at x=0.
What should be i, if we wish to obtain a force per unit length, just to overcome Earth's
acceleration, g=10 m/s?, for a payload of mass m=200 kg, given that D = 1 m? Hint: Find
the weight per unit length (mg/2D) and equate it to the right hand side of Eq.(2). You will
see that i, is huge, and this is why this technique is not appropriate to launch anything to
space, from Earth.
![Question 2
The rail gun is an electric device, conceived to launch a payload, installed on a rail system
composed of two parallel rails. Suppose you are located at the origin of the rails, and you look
ahead. A large electric current is sent along the left hand side (LHS) rail. A conducting fuse
element placed below the payload and in between the rails, conveys the current across, and
then back, along the right hand side (RHS) rail. The electric current sent in, produces a
magnetic field below the payload, and, the same current flowing through the magnetic field,
becomes subject to an upward magnetic force. The distance in between the rails is 2D.
Consider a two dimensional coordinate system (yOx): Let the Oy direction, the direction of
motion of the projectile through the launch, along the rails, and the Ox direction, the direction
perpendicular to Oy. Place the origin, in the middle of the rails, behind the payload.
Following the passage of the current of intensity i, the fuse element melts and vaporizes,
creating a gas, but still bearing the capability of conducting the electric current across the
rails, along the back side of the payload. The conducting gas, propels the projectile all along
the rails, and finally brings it to the stage of launch.
a
) Show that the net magnetic fieldintensity B(x) at x, created by both the LHS and the
RHS rail current, can be calculated to be
Ho iD
B(x) :
(1)
a(D? - x)
Precise the direction of the magnetic field vector B(x), and sketch B(x) versus x. Note that
the permeability constant, in MKS unit system, is u,=4n x 10-7.
Based (with the familiar notation), on the infinitely small magnetic force dF, exerted
on an infinitely small charge dq, moving with the velocity v, i.e. dF=v dq x B, show that
the magnitude dF(x), of the force exerted on the infinitely small piece of payload of width
|dx], around the location |x|, can be expressed as dF(x)= idx| B(x).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F35f75a04-2374-43e5-9e66-5970c48e4b4c%2F860cf77b-f427-4b61-946f-f002d656a594%2Fn55a3v_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Question 2
The rail gun is an electric device, conceived to launch a payload, installed on a rail system
composed of two parallel rails. Suppose you are located at the origin of the rails, and you look
ahead. A large electric current is sent along the left hand side (LHS) rail. A conducting fuse
element placed below the payload and in between the rails, conveys the current across, and
then back, along the right hand side (RHS) rail. The electric current sent in, produces a
magnetic field below the payload, and, the same current flowing through the magnetic field,
becomes subject to an upward magnetic force. The distance in between the rails is 2D.
Consider a two dimensional coordinate system (yOx): Let the Oy direction, the direction of
motion of the projectile through the launch, along the rails, and the Ox direction, the direction
perpendicular to Oy. Place the origin, in the middle of the rails, behind the payload.
Following the passage of the current of intensity i, the fuse element melts and vaporizes,
creating a gas, but still bearing the capability of conducting the electric current across the
rails, along the back side of the payload. The conducting gas, propels the projectile all along
the rails, and finally brings it to the stage of launch.
a
) Show that the net magnetic fieldintensity B(x) at x, created by both the LHS and the
RHS rail current, can be calculated to be
Ho iD
B(x) :
(1)
a(D? - x)
Precise the direction of the magnetic field vector B(x), and sketch B(x) versus x. Note that
the permeability constant, in MKS unit system, is u,=4n x 10-7.
Based (with the familiar notation), on the infinitely small magnetic force dF, exerted
on an infinitely small charge dq, moving with the velocity v, i.e. dF=v dq x B, show that
the magnitude dF(x), of the force exerted on the infinitely small piece of payload of width
|dx], around the location |x|, can be expressed as dF(x)= idx| B(x).
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