HELP PLEASE!!! 1. A 45-m length of wire is stretched horizontally between two ve
ID: 3278140 • Letter: H
Question
HELP PLEASE!!!
1. A 45-m length of wire is stretched horizontally between two vertical posts. The wire carries a current of 81 A and experiences a magnetic force of 0.15 N. Find the magnitude of the earth's magnetic field at the location of the wire, assuming the field makes an angle of 63.0° with respect to the wire.
T
2. Two insulated wires, each 2.90 m long, are taped together to form a two-wire unit that is 2.90 m long. One wire carries a current of 7.00 A; the other carries a smaller current I in the opposite direction. The two wire unit is placed at an angle of 58.0° relative to a magnetic field whose magnitude is 0.360 T. The magnitude of the net magnetic force experienced by the two-wire unit is 3.13 N. What is the current I?
A
3. A copper rod of length 0.78 m is lying on a frictionless table (see the drawing). Each end of the rod is attached to a fixed wire by an unstretched spring that has a spring constant of k = 76 N/m. A magnetic field with a strength of 0.15 T is oriented perpendicular to the surface of the table.
(a) What must be the direction of the current in the copper rod that causes the springs to stretch?
The current flows ---Select--- left-to-right right-to-left in the copper rod.
(b) If the current is 10 A, by how much does each spring stretch?
m
4. The drawing shows a thin, uniform rod that has a length of 0.35 m and a mass of 0.080 kg. This rod lies in the plane of the screen and is attached to the floor by a hinge at point P. A uniform magnetic field of 0.38 T is directed perpendicularly into the plane of the screen. There is a current I = 3.8 A in the rod, which does not rotate clockwise or counter-clockwise. Find the angle . (Hint: The magnetic force may be taken to act at the center of gravity.)
°
Explanation / Answer
Q1. given, length of wire l = 45 m
current in wire3, i = 81 A
magnetic force on the wire, F = 0.15 N
angle of earths magentic field with the direction of current = theta = 63 degree
now, for ceon a current carrying conductor in a magnetic field is F = Bilsin(theta)
so, appying the formula
0.15 = B*81*45*sin(63)
so, B = 4.61*10^-5 T
Q2. so force on the fused conductor would be
F = B(I - i)lsin(theta)
B = 0.36 T
I = 7 A
l = 2.9 m
F = 3.13 N
theta = 58 deg
so, 3.13 = 0.36*(7 -i)2.9sin(58)
so i = 3.467 A
Q3. a. For downward force to exert, the current must be left to right ( from left hand rule)
b. F = Bil = 0.15*10*0.78 = F = 2kx = 2*76*x
x = 0.00769 m extension in both the springs
Q4. given,. length, l = 0.35 m
mass, M = 0.08 kg
let emagnetic force be F
F = Bil = mg*cos(theta)
0.38*3.8*0.35 = 0.08*9.81*cos(theta)
theta = 49.910 degree
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