Phys112 Fall 2014 A single square loop of wire (20 cm times 20 cm) is free to mo

Question

Phys112 Fall 2014 A single square loop of wire (20 cm times 20 cm) is free to move and carries a current of 1.2 A in a counterclockwise direction when viewed from above. The loop face is at an angle of 25degree to the horizontal and it sits in a 0.8 T uniform magnetic field that points directly DOWN. What is the FORCE on the right edge (*) of the loop due to the magnetic field? Zero 0.19 N to the left 0.08 N to the right 0.25 N down 1.8 N up What is the magnitude of the TORQUE on the loop and how does it ROTATE? (give the direction of rotation when viewed from the side - i.e. looking directly at the figure drawn). 0.035 Nm, counterclockwise 0.016 Nm, counterclockwise 0.035 Nm, clockwise 0.016 Nm, clockwise NONE of the above. A circuit consists of a circular loop of wire (radius = 60 cm) connected to a 270 2 resistor. The circuit sits in a uniform 2.0 T magnetic field that points into the page (perpendicular to the loop area). The magnetic field is steadily decreased at a rate of 0.34 T/s. While the field is changing, what CURRENT (if any) is INDUCED in the circuit? (The wire has negligible resistance) Zero, no current is induced 1.4 mA 8.6 mA 23 mA 100 mA

Explanation / Answer

55)
Force = B x I x L
   = 0.8 T x 1.2 A x 0.2 m = 0.192 N.
Using Fleming's left hand rule, thumb indicates the direction of force, which is pointing to the left.

56)
Force on the side facing us and the side opposite to that will cancel each other (the forces on them are equal and outwards)
Torque = cross product of radial vector and the force.
= r x F, r = 0.1 m
   = r F sin(25) = 0.1 x 0.192 x 0.423 = 0.008 Nm
Total torque = 2 x 0.008 = 0.016 N m, direction is clockwise

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