In the figure below, a metal wire of mass m 30.0 mg can slide with negligible fr
ID: 1391677 • Letter: I
Question
In the figure below, a metal wire of mass m 30.0 mg can slide with negligible friction on two horizontal parallel rails separated by a distance d 2.6 cm The track lies in a vertical uniform magnetic field of magnitude 63.0 mT. At time t o, a function generator device G is connected to the rails, producing a constant current I 310.0 mA in the wire and rails (even the wire moves) (a) Calculate the magnitude and direction of the force the magnetic field exerts on the wire segment. (b) Determine the wire segment acceleration (magnitude and direction). (c) At t -60.0 ms, what is the wire velocity (magnitude and direction).Explanation / Answer
Given that,
mass of the wire = m = 30 mg = 30 x 10-3 kg ; d = 2.6 cm = 0.026 cm;
B = 63 mT = 63 x 10-3 T; I = 10 mA = 0.01 A
(a)The Force that the B field exerts on the wire will be given by;
F = B I L
In our case, L = d, So F = B I d
F = 63 x 10-3 x 0.01 x 0.026 = 1.64 x 10-5 T- A - m (Newtons)
Hence, F = 1.64 x 10-5 Newtons,
This force will be perpendicular to B field, so its direction will be towards left.
(b)We know that, F(magnetic) = B I d and also F = ma. These two forces must be balancing each other.
F(magnetic) = F
B I d = m a => a = B I d / m = 1.64 x 10-5 / 0.03 = 5.46 x 10-4 m/s2
Hence, a = 5.46 x 10-4 m/s2, directed to left
(c)given t = 60 ms = 0.06 sec
We know that,
v = u + at (but u = 0, in our case)
v = at = 5.46 x 10-4 x 0.06 = 3.28 x 10-5 m/s
Hence, v = 3.28 x 10-5 m/s,directed to left