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One end of a resistanceless conducting horizontal pair of rails having a distance L between them, is connected to a capacitor of capacitance C that has been charged to a voltage V0. The inductance of the assembly is negligibe. The system is placed in a uniform, vertical magnetic field B (the shaded region in the figure). A frictionless conducting rod of mass m and resistance R is placed perpendicularly onto the track. The polarity of the capacitor is such that the rod is repelled from the capacitor when the switch is turned on.


i) What is the maximum velocity of the rod?


ii) Under what conditions is the efficiency of this "electromagnetic gun" maximum?

Amit Saxena , 12 Years ago
Grade upto college level
anser 1 Answers
Askiitians Tutor Team

To tackle this problem, we need to analyze the dynamics of the rod in the context of electromagnetic principles. The setup you've described resembles a type of electromagnetic launcher, often referred to as a railgun. Let's break down the two parts of your question step by step.

Determining the Maximum Velocity of the Rod

When the switch is turned on, the charged capacitor discharges through the rod, creating a current. This current interacts with the magnetic field, generating a force that propels the rod. The force \( F \) acting on the rod can be described by the Lorentz force law:

F = I \cdot L \cdot B

Here, \( I \) is the current flowing through the rod, \( L \) is the length of the rod, and \( B \) is the magnetic field strength. The current \( I \) can be expressed in terms of the capacitor's voltage \( V_0 \) and the resistance \( R \) of the rod:

I = \frac{V_0}{R}

Substituting this into the force equation gives:

F = \frac{V_0}{R} \cdot L \cdot B

According to Newton's second law, the force acting on the rod also equals the mass of the rod times its acceleration:

F = m \cdot a

Setting these two expressions for force equal to each other, we have:

m \cdot a = \frac{V_0}{R} \cdot L \cdot B

To find the maximum velocity, we can integrate the acceleration over time. However, for simplicity, we can use the work-energy principle. The work done on the rod as it moves a distance \( d \) is equal to the change in kinetic energy:

W = F \cdot d = \frac{1}{2} m v^2

Substituting the expression for force:

\frac{V_0}{R} \cdot L \cdot B \cdot d = \frac{1}{2} m v^2

Solving for \( v \), we find:

v = \sqrt{\frac{2 \cdot V_0 \cdot L \cdot B \cdot d}{R \cdot m}}

This equation gives us the maximum velocity of the rod as it is propelled by the electromagnetic force generated by the capacitor discharge.

Maximizing Efficiency of the Electromagnetic Gun

Efficiency in this context can be defined as the ratio of the useful work output (kinetic energy of the rod) to the energy input from the capacitor. To maximize efficiency, we need to consider several factors:

  • Minimizing Resistance: The resistance \( R \) of the rod should be as low as possible. This reduces energy losses due to heat, allowing more energy to be converted into kinetic energy.
  • Maximizing Voltage: A higher initial voltage \( V_0 \) from the capacitor increases the current and thus the force acting on the rod, leading to greater acceleration and velocity.
  • Optimizing Magnetic Field Strength: A stronger magnetic field \( B \) enhances the force acting on the rod, contributing to higher velocities.
  • Length of the Rod: Increasing the length \( L \) of the rod can also increase the force, provided it remains within the physical constraints of the system.

In summary, the efficiency of the electromagnetic gun is maximized when the resistance is minimized, the voltage is maximized, and the magnetic field strength is optimized. By carefully balancing these factors, one can achieve the best performance from the system.

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