heeeelp 27-50 coil is oriented parallel to the x-axis, a force of 1.20 n applied
ID: 3898212 • Letter: H
Question
heeeelp
27-50
coil is oriented parallel to the x-axis, a force of 1.20 n applied to the edge of the coil in the positive y-direction can keep it from rotating. Calculate the strength of the magnetic field. Twenty loops of wire are tightly wound around a round pencil that has a diameter of 6.00 mm. The pencil is then placed in a uniform 5.00-T magnetic field, as shown in the figure. If a 3.00-A current is present in the coil of wire, what is the magnitude of the torque on the pencil? A copper wire with density rho = 8960 kg/m3 is formed into a circular loop of radius 50.0 cm. The cross-sectional area of the wire is 1.00 10-5 m2, and a potential difference of 0-0120 V is applied to the wire. What is the maximum angular acceleration of the loop when it is placed in a magnetic field of magnitude 0.250 T? The loop rotates about an axis in the plane of the loopExplanation / Answer
The magnetic moment of the solenoid is
m = I*N*S =3*20*pi*(3*10^-3)^2=1696.5*10^-6 A*m^2
The direction of magnetic moment vector is perpendicular to the loop plane.
The torque of the external magnetic field is
T = m x B = m*B*sin(alpha) =1696.5*10^-6*5*sin(60) = 7345.9*10^-6 N*m
(x is the cross product)
(http://en.wikipedia.org/wiki/Magnetic_moment#Definition)
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