Question

The figure below shows a bar of mass m = 0.280 kg that can slide without friction on a pair of rails separated by a distance ℓ = 1.20 m and located on an inclined plane that makes an angle θ = 29.5° with respect to the ground. The resistance of the resistor is R = 1.50 Ω, and a uniform magnetic field of magnitude B = 0.500 T is directed downward, perpendicular to the ground, over the entire region through which the bar moves. With what constant speed v does the bar slide along the rails?

Answer 1

Here is the free body diagram,

Here, In x - direction, we can write

F - Nsin

F = Nsin -------------- (1)

In y - direction, we can write

mg - Ncos

mg = Ncos ------------------- (2)

Dividing (1) by (2)

F / mg = tan

(Note - sin / cos = tan )

F = mgtan

But we also know that

F = ILB

so,

ILB = mgtan

I = mgtan / LB

induced emf = IR

induced emf = (mgtan / LB) * R ------------- (3)

but also, we know that

induced emf = vBLcos -------------- (4)

combining (3) and (4), we get

vBLcos = (mgtan / LB) * R

v =  (mgtan / L²B² cos) * R

Now, we have got the equation, just put in the values,

v = (0.280 * 9.8 * tan 29.5 / 1.20²0.5² cos 29.5) * 1.50

v = 7.4322 m/s