A straight, horizontal rod slides along a parallel conducting rails at an angle

Question

A straight, horizontal rod slides along a parallel conducting rails at an angle an angle with the horizontal of 34 degree. The rails are connected at the bottom by a horizontal rail so that the rod and rails forms a closed rectangular loop. A uniform vertical field exists throughout the region. Assume the rod remains in contact with the rails as it slides down the rails. The rod experiences no friction or air drag. The rails and rod have negligible resistance. The acceleration due to gravity is 9.8 m/s^2. If the velocity of the rod is 1.5 m/s, what is the current through the resistor? Answer in units of mA. B). what is the terminal velocity of the rod? Answer in units of m/s.

Explanation / Answer

A).Using Faraday's law ad noting that the distance of therails from the rod changes with time at a rate (dx/dt) = v,theinduced motional emf is E = -(dB/dt) = -(d/dt)(B * l * v) = -B *l * (dx/dt) or E = -B * l * v Because the resistance of the circuit is R,the magnitude ofthe induced current is I = |E|/R = (B * l * v/R) B is the magnetic field,l is the length of the rod,v is thespeed of the rod and R is the resistance. B).Applying Newton's second law to the rod in the horizontaldirection Fx = ma = m * (dv/dt) = -I * l * B Substituting I = (B * l * v/R) we get m * (dv/dt) = -(B2 * l2/R) * v or (dv/v) = -(B2 * l2/mR) * dt Integrating the above equation on both sides we get ln(v/vi) = -(B2 * l2/mR) *t or v = vi * e-t/ where = (m * R/B2 * l2) the velocity of the rod is vi = 1.5 m/s m is the mass of the rod where = (m * R/B2 * l2) the velocity of the rod is vi = 1.5 m/s m is the mass of the rod

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