1.The coil of wire in the drawing is a right triangle and is free to rotate abou
ID: 2280092 • Letter: 1
Question
1.The coil of wire in the drawing is a right triangle and is free to rotate about an axis that is attached along side AC. The current in the loop is I = 5.39 A, and the magnetic field (parallel to the plane of the loop and side AB) is B = 1.65 T.
The coil of wire in the drawing is a right triangle and is free to rotate about an axis that is attached along side AC. The current in the loop is I = 5.39 A, and the magnetic field (parallel to the plane of the loop and side AB) is B = 1.65 T. What is the magnetic moment of the loop? What is the magnitude of the net torque exerted on the loop by the magnetic field? Two long, straight wires are separated by 0.12 m. The wires carry currents of 1.0 A in opposite directions, as the drawing indicates. Find the magnitude of the net magnetic field at the point A. Find the magnitude of the net magnetic field at the point B. A -6.00-mu C charge is moving with a speed of 6.70 times 104 m/s parallel to a very long, straight wire. The wire is 5.00 cm from the charge and carries a current of 77.0A in a direction opposite to that of the moving charge. Find the magnitude and direction of the force on the charge.Explanation / Answer
1.
Height of the traingle
tan55 =h/2
h=2.856 m
Area
A=(1/2)*h*b =(1/2)*2.856*2
A=2.856 m^2
a)
magnetic moment
H=I*A =5.39*2.856
H=15.4 A-m^2
b)
T=B*H=1.65*15.4
T=25.4 N-m
2)
r=0.12 m
d=0.03 m
x=0.06
a)
Magnetic field at point A is
B=(uoI/2pi)[1/X -1/(r+d)]
B =(uoI/2pi)[1/0.06 -1/(0.03+0.12)]
B=[(4pi*10^-7)*1/2pi]*10
B=2*10^-6 T
b)
Magnetic field at point B is
B=2uo*I/pi*r
B=2*(4pi*10^-7)*1/pi*0.12
B=3.33*10^-6 T
3)
B=Uo*I/2pir
B=(4pi*10^-7)*77/2pi*0.05
B=3.08*10^-4 T
F=qVB =(6*10^-6)(6.7*10^4)*3.08*10^-4
F=0.1238 mN
4)
r=mV/qB
here V,q and B are constants so
rp/re =mp/me =(1.67*10^-27)/(9.11*10^-31)
rp/re =1833.15
5)
F =(uo*Iw*IL*L1/2pi)[1/d -1/(d+L2)
F=[(4pi*10^-7)*15*28*0.5/2pi]*[1/0.11 -1/(0.11+0.15)]
F=2.2*10^-4 N
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