physics problem : Faraday's law?
Figure
http://www.bluemelon.com/photo/16392/714857-T80060...
shows a bar of mass m = 0.2
kg that can slide without friction on a pair of rails
separated by a distance l =1.2 m and located on an
incline plane that makes an angle = 25.0° with respect
to the ground. The resistance of the resistor is R = 1.0
and a uniform magnetic field of magnitude of B =
0.5 T is directed downward, perpendicular to the
ground, over the entire region through which the bar
moves. With what speed v does the bar slide along the
rails?
Comments
Component of B perpendicular to the plane
= Bcos25°
=> emf induced in the coil
ε = Bcos25° * 1.2 * v =
=> current in the bar,
I = ε/R = (v/R) * 1.2 Bcos25°
Force opposing the downward motion
F = I * L * Bcos25° = (v/R) * (1.2 Bcos25°)^2
As the bar starts sliding, v keeps on increasing and so does F till F balances the downward component of the weight of the bar
=> mgsin25° = (v/R) * (1.2 Bcos25°)^2
=> velocity, v
= (mgRsin25°) / (1.2 Bcos25°)^2
= (0.2 * 9.81 * 1.0 * sin25°) / (1.2 * 0.5 * cos25°)^2
= (0.8292) / (0.2957)
= 2.80 m/s.
Hello mmmeee, let me derive the right expression to get the required velociy.
If I be the current through the rod of length l to whose length B cos 25 magnetic field induction is acting then the force on the rod up the plane will be B cos 25 l sin 25.
To get the current we need the emf induced in the rod which is given by B cos 25 l v. Let v be the required velocity to be found out.
Hence the current in the rod I = e/R = Blv cos 25 / 1 = Blv cos 25.
So the force upward the inclined plane, B^2 l^2 cos^2 25 * sin 25 v
But due to inclined position the sine component of the weight of the rod would act down the inclined plane. That would be mgsin 25.
Hence the net force has to be zero as we assume the rod is sliding down with uniform speed v.
Hence B^2 l^2 cos^2 25 * sin 25* v = mg sin 25
So the required v = mg / B^2 l^2 cos^ 25 (cancellig sin 25 on both sides)
Please plug the values for m, g, B, l and get the value of v.
since its a constant speed Fg = Fb
gravitational force equals magnetic force
Fg = mg sin25
EMF = dΦ/dt = d/dt (Blx cos25) = B cos25 lv
I = Blv cos25/R
Fb = BIL cos25 = B^2 l^2 v (cos25)^2/R
Fb = Fg
B^2 l^2 v cos25/R = mg sin25
v = (R mg sin25)/(B^2 l^2 (cos25)^2)
v = (1 * 0.2 * 9.8 * sin 25)/(0.25 * 1.44 (cos25)^2) = 2.80 m/s
Flux = B cos 25 * area; emf, e = rate of change of flux; current,i = e/r and force = Bil.
{[B*l*(cos 25)*v*dt]/Rdt}*B*(cos 25)*l = (mg)*(sin 25)
v = [(m*g)*(tan 25)*R]/[(B^2)*(cos 25)*(l^2)]
= [0.2*9.8*(tan 25)*1]/[(0.5^2)*0.9063*(1.2)^2] = 2.80 m/s