A straight wire (length L = 17.7 cm, mass m = 71 g) is part of an electric circu

Question

A straight wire (lengthL=17.7cm, massm=71g) is part of an electric circuit that includes a battery and a switch (initially open). The wire lies on a table, oriented so it runs north-south,28.3m below the ceiling. In the entire region there is a uniform magnetic fieldB=0.125T runnning east-to-west.

a) The switch is closed, and currentI=7.49runs through the wire from south-to-north. Find the normal force exerted by the table on the wire.
N=N
b) Suppose the current in the wire is suddenly increased to178.81A. Neglecting friction, find the time after the current is turned up that the wire would hit the ceiling.
t=s

c) With the swtich open (soI= 0), and the wire is turned so it lies on the table at an angle?=39.8oeast of north. Now the switch is closed, and178.81A flows through the wire. Neglecting friction, find the time after the current is turned up that the wire would hit the ceiling.
t=s

Explanation / Answer

(F_{mag} = IB imes l = IBLsin heta = 7.49 imes 0.125 imes 0.177 = 0.165N)

since the F is downward

(N_{table} = mg + F_{mag} = 71 imes 10^{-3} imes 9.8 + 0.165 = 0.862N)

b)current is turened up in north to south direction = 178.81A

again the force due t o magnetic field is

(178.81 imes 0.125 imes 0.177 = 3.956N)

since the current is reversed the force will be upward

so F_net upward = F_mag-mg = 3.260

acceleration a = 3.260/0.071 = 45.92m/s^2.

by eq of motion

(S = ut + at^2/2)

since u= 0

(t = sqrt{2S/a} = sqrt{2 imes 28.3/45.92} = 1.11s)

c) since the wire is making the angele 39.8 with north to south, the angle between mag feild and the direction of th current is 90-39.8 = 50.2degree.

so (F_{mag} = 178.81 imes 0.125 imes 0.177 imes sin50.2 = 3.039N)

so the F_{net} = 3.039-mg = 2.34N

a= 2.34/0.071 = 33.01m/s^2

by the same eq of motion

(t = sqrt{2 imes 28.3/33.01 } = 1.30s)

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