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22 Nov 2008 00:21:39 IST

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A primitive electromagnetic gun.

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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?

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Comments (7)

Karthik M

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22 Nov 2008 15:17:21 IST

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When the switch is thrown to the other side, the capacitor discharges through the resistor, ie, the rod.

The transient voltage across the rod at any time t = $V_0 e^{-\frac{t}{\tau}}$

Thus, i(t) = $\frac{V_0 e^{-\frac{t}{\tau}}}{R}$

Now, $m\frac{dv}{dt}= BL\frac{V_0 e^{-\frac{t}{\tau}}}{R}$

$\int^v_0 m \, dv = \int^t_0 BL\frac{V_0 e^{-\frac{t}{\tau}}}{R} \, dt$

$v(t) = \frac{BV_o CL}{m}(1- e^{-\frac{t}{\tau}})$

Thus, the maximum value is BVoCL/m.

For the second part, CV_o ^2 = mv^2 + 2i^2 R t by COE. So Differentiating and equating to 0 should give the condition (Output is mv²)

Any flaws?

Karthik M

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22 Nov 2008 15:46:18 IST

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For the second part, $\frac{d}{dt} \int^t_0 i(t) ^2 R dt$ = 0, for minimum losses. Chuck what I said in my previous post.

Anant Kumar

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22 Nov 2008 17:01:59 IST

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There is one serious flaw. In fact, your starting point is itself not correct. This is not a case of pure RC discharge. The motion emf is also going to affect it which you have neglected.

Karthik M

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22 Nov 2008 17:44:11 IST

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Hmm... not quite sure by what you mean that this is not a case of pure RC discharge, but is the current at any time
i = V(t) - BLv/R ?

Anant Kumar

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22 Nov 2008 18:10:59 IST

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yes now you are on the right track..

Karthik M

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22 Nov 2008 20:30:57 IST

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Yeah, that was a grave error. Anyway,

$m\frac{dv}{dt} = \frac{BL}{R} (V_o e^{-\frac{t}{\tau}} - BLv)$

I guess I forgot my 12th math. Not able to solve this DE so far :(

Dipanjan

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22 Nov 2008 22:54:45 IST

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Re:A primitive electromagnetic gun.

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