A coil of wire is mounted between two magnets. When there is a current in the coil,it rotates.The diagram shows the position of the coil, viewed from one end, when it is at an angleof 20° to the horizontal. The current is into the page at X and out of the page at Y.There are 10 turns on the coil and the current in the coil is 6.9 A.Determine the resultant moment about P of the magnetic forces acting on the coil.length of coil = 5.0 cmwidth of coil = 3.5 cmcurrent = 6.9 Anumber of turns = 10magnetic flux density = 0.07 T(4)...............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................Moment about P = ......................................................(Total for Question 11 = 4 marks)
Question
A coil of wire is mounted between two magnets. When there is a current in the coil,it rotates.The diagram shows the position of the coil, viewed from one end, when it is at an angleof 20° to the horizontal. The current is into the page at X and out of the page at Y.There are 10 turns on the coil and the current in the coil is 6.9 A.Determine the resultant moment about P of the magnetic forces acting on the coil.length of coil = 5.0 cmwidth of coil = 3.5 cmcurrent = 6.9 Anumber of turns = 10magnetic flux density = 0.07 T(4)...............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................Moment about P = ......................................................(Total for Question 11 = 4 marks)
Solution
To determine the resultant moment about point P of the magnetic forces acting on the coil, we can use the formula:
Moment = N * B * A * sin(theta)
Where: N = number of turns = 10 B = magnetic flux density = 0.07 T A = area of the coil = length * width = 5.0 cm * 3.5 cm = 17.5 cm^2 = 0.00175 m^2 theta = angle between the magnetic field and the normal to the coil = 20°
Plugging in the values, we have:
Moment = 10 * 0.07 T * 0.00175 m^2 * sin(20°)
Calculating this expression, we find:
Moment = 10 * 0.07 * 0.00175 * sin(20°) = 0.000245 Nm
Therefore, the resultant moment about point P of the magnetic forces acting on the coil is 0.000245 Nm.
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