A conducting bar with mass m and length L slides over...

A conducting bar with mass $m$ and length $L$ slides over horizontal rails that are connected to a voltage source. The voltage source maintains a constant current $I$ in the rails and bar, and a constant, uniform, vertical magnetic field $\vec{B}$ fills the region between the rails (Fig. $\mathbf{P 2 7 . 5 9}$ ). (a) Find the magnitude and direction of the net force on the conducting bar. Ignore friction, air resistance, and electrical resistance. (b) If the bar has mass $m,$ find the distance $d$ that the bar must move along the rails from rest to attain speed $v$. (c) It has been suggested that rail guns based on this principle could accelerate payloads into earth orbit or beyond. Find the distance the bar must travel along the rails if it is to reach the escape speed for the earth $(11.2 \mathrm{~km} / \mathrm{s}) .$ Let $B=0.80 \mathrm{~T}, I=2.0 \times 10^{3} \mathrm{~A}, m=25 \mathrm{~kg}$ and $L=50 \mathrm{~cm} .$ For simplicity assume the net force on the object is equal to the magnetic force, as in parts (a) and (b), even though gravity plays an important role in an actual launch in space.

A conducting bar with mass $m$ and length $L$ slides over horizontal rails that are connected to a voltage source. The voltage source maintains a constant current $I$ in the rails and bar, and a constant, uniform, vertical magnetic field $\vec{B}$ fills the region between the rails (Fig. $\mathbf{P 2 7 . 5 9}$ ). (a) Find the magnitude and direction of the net force on the conducting bar. Ignore friction, air resistance, and electrical resistance. (b) If the bar has mass $m,$ find the distance $d$ that the bar must move along the rails from rest to attain speed $v$. (c) It has been suggested that rail guns based on this principle could accelerate payloads into earth orbit or beyond. Find the distance the bar must travel along the rails if it is to reach the escape speed for the earth $(11.2 \mathrm{~km} / \mathrm{s}) .$ Let $B=0.80 \mathrm{~T}, I=2.0 \times 10^{3} \mathrm{~A}, m=25 \mathrm{~kg}$
and $L=50 \mathrm{~cm} .$ For simplicity assume the net force on the object is equal to the magnetic force, as in parts (a) and (b), even though gravity plays an important role in an actual launch in space.

Key Concepts

Magnetic Force on a Current-Carrying Conductor

This concept describes the force experienced by a conductor when it carries an electric current in the presence of a magnetic field. The force is given by the vector product of the current element and the magnetic field, typically expressed as F = I L x B. This principle is central to devices like railguns and motors, where the interaction between current and a magnetic field produces mechanical motion.

Right-Hand Rule for Determining Force Direction

The right-hand rule is a mnemonic for determining the direction of the force on a current-carrying conductor placed in a magnetic field. By aligning the thumb, index, and middle finger perpendicularly (with the thumb representing the force, the index finger the current direction, and the middle finger the magnetic field direction), one can correctly predict the orientation of the force. This rule emphasizes the vector nature of the magnetic force resulting from the cross product involved in its calculation.

Work-Energy Principle

The work-energy principle states that the work done on an object by a net force is equal to the change in its kinetic energy. In the context of a moving conductor under the influence of a constant magnetic force, this principle is used to equate the product of the force and the displacement to the kinetic energy acquired by the conductor. This allows for the calculation of displacement required to reach a particular speed.

Kinematics Under Constant Acceleration

Kinematics for objects under constant acceleration provides relationships between displacement, velocity, and time. When a constant magnetic force results in constant acceleration, equations from kinematics can be applied to compute the distance traveled to reach a specified velocity. This approach is fundamental in problems where energy is imparted steadily over a known distance.

Constant Current and Circuit Dynamics

In systems where a voltage source maintains a constant current, the current remains fixed even as the physical configuration of the circuit changes. This assumption simplifies the analysis of electromagnetic forces by ensuring that the magnetic force remains constant provided the other parameters (like the magnetic field and conductor length) are unchanged. The interplay between the voltage source and the conductor’s motion is crucial in applications like railguns, where consistent acceleration is desired.

Transcript

00:01 Here we're going to examine the operation of a so -called rail gun.

00:08 And the principle of operation involves the lorentz force on a moving charge.

00:17 And in this case, we have a bar.

00:23 So the force is going to translate into i -l cross -es.

00:33 B, where l is the length of the bar, i is the current through the bar, and the magnetic field is hopefully the magnetic field in which the bar is sitting.

00:50 So let's take a look.

00:51 The current is being produced by a voltage source, and the current is maintained by an external circuit over which supposedly the bar rides with zero resistance.

01:06 If we use the right -hand rule, we can figure out which direction the force acts on that bar.

01:13 And what i like to do with my right -hand rule is i like to point the fingers on my right -hand in the direction of the velocity, or in this case, the current, and then let them naturally curl in the direction of the magnetic field, which in this case is out of the page.

01:35 And then my thumb will give the direction of the force.

01:40 So that's my palm sticking out.

01:43 And so the force is in the direction propelling this bar away from the voltage source.

01:57 So the idea then is to figure out how much kinetic energy you can give to this bar.

02:06 There's a couple ways to do this.

02:09 The easiest way i believe is to use newton's second law, as this is a kinematics type situation.

02:18 Newton's second law says that the sum of the forces on an object equals its mass times its acceleration.

02:33 Here, we only have one force.

02:36 It's in the x direction, and the current and the b are at right angle.

02:44 So we can simply write the magnitude down of that x force as ilb, and that should equal m -a -sup -x...

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