A metallic rod (negligible resistance) can slide frictionlessly on two parallel metallic rails separated by a distance l in the presence of a perpendicular magnetic field (B). The ends of the rails are joined through a source of emf E and an inductance L in series. Obtain a differential equation for the velocity of the rod and show that the velocity varies simple harmonically about a mean value with a time period T. Obtain expressions for velocity and T. My try: Net emf in circuit E–Ldi/dt. Now i pull the rod by a distance dx so emf generated would be Bldx/dt , now dx/dt is v(t) so induced emf is Blv(t) . Now net emf in circuit would be E–Ldi/dt±Blv(t) (±as o don't know what direction B points), but now I can't think of anything to find the solution . It's given resistance less so F=ilB won't work and I don't know what can I do more . Please help me in trying out the solution.
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"Now i pull the rod by a distance dx [...]" Pulling the rod was not mentioned in the question. Nor is pulling necessary. As soon as there is current in the rod it will experience a motor effect (Laplace) force of magnitude $F=Bil$ parallel to the rails. This gives you the acceleration.
"E–Ldi/dt±Blv(t)" This is indeed the net emf in the circuit and it is zero because there is no resistance and no potential drop. Use Lenz's law to figure out what sign is needed.
You now have two equations, and SHM emerges quite nicely.
answered Aug 18, 2021 at 17:00
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$\begingroup$ Thanks a lot @Philip Wood i understood it and was able to solve it. $\endgroup$– User 1Commented Aug 18, 2021 at 17:49
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$\begingroup$ Good. In the UK the $F=Bil \sin \theta$ force is usually called the motor effect force. As a matter of interest what do you call it? $\endgroup$ Commented Aug 18, 2021 at 17:59
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$\begingroup$ We generally call it Lorentz force on conductor . $\endgroup$– User 1Commented Aug 19, 2021 at 6:07
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$\begingroup$ Thank you. The definition of the Lorentz force is the force acting on a charged particle in an electric and/or magnetic field, so $\vec F_\text {Lorentz} =q(\vec E + \vec v \times \vec B)$. The second term on the right is the magnetic part of the LF. The force that acts on a current-carrying wire arises from the magnetic LFs on the moving free electrons, but shouldn't, strictly, be called the LF. [The LF doesn't act directly on the wire itself.] There seems to be no consensus on what to call the force on the wire! $\endgroup$ Commented Aug 19, 2021 at 8:18