A non-ideal conducting bar of mass m = 0.27kg, which carries a resistance of R =

Question

A non-ideal conducting bar of mass m = 0.27kg, which carries a resistance of R = 1.50 ohm slides without friction on two perfectly conducting rails that are connected at one end by a power supply as illustrated in the above diagram. The two rails are separated by a distance l = 72.0cm. A magnetic field is perpendicular to the plane that the bar and rails make and has a magnitude of |B| = 1.60T. The bar is allowed to slide along the vertical direction in a gravitational field, (a) If the initial velocity is zero, then What must the potential difference across the power supply be for the bar to stay at its current location (hint: chapter 29 material)? (b) Let us assume that we remove the power supply. The total force on the object will be Ftot = Fgravity - Fmag, and therefore, m(dv/dt) = mg -I l B. Substituting IR = B l v into the differential equation gives m(dv/dt) = mg - B2l2, 2v/R, which is an inhomogeneous differential equation. Upon solving this equation for the bar initially at rest, we find that v = g tau (1-e-t/tau), where t = mR/B2 l 2. What is the maximum velocity that the bar can achieve?

Explanation / Answer

a)potential =mg/(BLR)= 1.53 V

b)for maximum velocity dv/dt=0

vmax=mgR/[(B^2 )*(L^2)]=2.993 m/sec

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