1) A single circular loop of wire of radius 0.75m carries a constant current of
ID: 1683937 • Letter: 1
Question
1) A single circular loop of wire of radius 0.75m carries a constant current of 3.0A. The loop may be rotated about an axis that passes through the center and lies in the plane of the loop. When the orientation of the normal to the loop with respect to the direction of the magnetic field is 25 degrees, the torque on the coil is 1.8 N.m. What is the magnitude of the uniform magnetic field exerting this torque on the loop? 2) The Earths magnetic field passes through a square tabletop with a magnitude of 4.95x10^-5 T and is directed at an angle of 165 degrees relative to the normal of the tabletop. If the tabletop has 1.50 m sides, what is the magnitude of the magnetic flux through it? 3) An overhead electric power line carries a maximum current of 125 A. What is the magnitude of the maximum magnetic field at a point 4.50 m directly below the power line?1) A single circular loop of wire of radius 0.75m carries a constant current of 3.0A. The loop may be rotated about an axis that passes through the center and lies in the plane of the loop. When the orientation of the normal to the loop with respect to the direction of the magnetic field is 25 degrees, the torque on the coil is 1.8 N.m. What is the magnitude of the uniform magnetic field exerting this torque on the loop? 2) The Earths magnetic field passes through a square tabletop with a magnitude of 4.95x10^-5 T and is directed at an angle of 165 degrees relative to the normal of the tabletop. If the tabletop has 1.50 m sides, what is the magnitude of the magnetic flux through it? 3) An overhead electric power line carries a maximum current of 125 A. What is the magnitude of the maximum magnetic field at a point 4.50 m directly below the power line?
Explanation / Answer
Number of turns in a circular loop is N = 1 current in the loop is I = 3.0 A radius of the wire is r = 0.75 m Area A = p r^2 = p ( 0.75 m) ^2 = 1.76 m^2 torque t = 1.8 N m t = N A I B sin? 1.8 N m = 1 * 1.76 * 3.0 * B * sin ( 25 ) magnetic field B = 0.80 T b ) magnitude of a magnetic field on a square table top is B = 4.95*10^-5 T side of the square table is a = 1.50 m Area A = a^2 = 1.50*1.50 = 2.25 m^2 magnetic flux is f _B = B A cos ? = 4.95*10^-5 T * 2.25 * cos (165-90 ) = 2.88*10^-5 Weber c ) maximum current I = 125 A at a distance R = 4.50 m magnetic field B = µ_0 I / 2p R = 4p*10^-7 *125 A / 2p * 4.50 m = 55.5* 10^-7 TRelated Questions
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