Sliding + Oscillating Rod in a Circuit!

Side rail of length 2 L 2L are fixed on a horizontal plane at a distance from each other. These ends are connected by two identical ideal batteries with emf E E by resistance less wires (see figure). On the rails is a rod of mass m m , which may slide along them. The entire system is placed in a uniform vertical magnetic field B B . Assuming that the resistance of the rod is R R and the resistance per unit length of each of the rails equal to ρ ρ , find the period of small oscillations (in sec.) arising from shifting the rod from the equilibrium along the rails. Neglect friction, internal resistance of batteries and induced emf in the rod. [Take : B = π T B = \pi T , ε = π v o l t ε = π volt , = 0.5 m ℓ = 0.5m , L = 1 m L = 1m , ρ = 1 Ω / m ρ = 1Ω / m , R = 0.25 Ω R = 0.25Ω , m = 100 g m m = 100 gm ]


The answer is 1.00.

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2 solutions

Nishant Rai
May 15, 2015

@Nishant Rai Can you tell me where I went wrong.. .. .. .. .. .. .. .. .. ..
Here is the solutuon

raj abhinav - 1 year, 2 months ago
Ace Pilot
Sep 4, 2015

Easy prob. Initially whole system is in equilibrium When u displace the slider by x<<l, a current i i=Epx/R+pL is established. ( by combinations of cell). Therfore force on slider is ilB As F proportional to x, it will perform SHM Therefore T=2π\w 2π√[(R+pL)/EplB] Substitute values nd will get d ans 1

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