Part A Calculate the magnitude of the torque due to the magnetic force on the ro

ID: 1408222 • Letter: P

Question

Part A

Calculate the magnitude of the torque due to the magnetic force on the rod, for an axis at P.

Part B

Calculate its direction. (Counterclockwise or Clockwise)

Part C

Is it correct to take the total magnetic force to act at the center of gravity of the rod when calculating the torque? (Yes or No)

Part E

When the rod is in equilibrium and makes an angle of 53.0 ? with the floor, is the spring stretched or compressed? (Stretched or Compressed)

Part F

How much energy is stored in the spring when the rod is in equilibrium?

53.0°× ×

here,
spring constant, k = 4.80 N/m
Magnatic field, b = 0.330 T
length of rod, l = 0.20 m
Current, i = 6.50 A

Part A:
magnatic force, F = I*B*dr

since force is perpendicular to rod , therefore torque dt is given as,
magnatic torque, dt = I*B*r*dr

upon integration from 0 to L we get,

Torque, t = 0.5*I*L^2*B

t = 0.5 * 6.50 * 0.2^2 * 0.330
t = 0.0429 N.m

Part B:
Since torque is positive so it is acting in clockwise direction

Part C:
no, since it is higed so net force will be maximum when applied to end point

Part E:
Since, rotational torque, tr = 0.5*I*L^2*B
also translational torque, tt = k*x*L*SinA

rewriting eqn for compressed or stretched distance, x = I*L*B/(2*k*SinA)
x = 6.50*0.2*0.330/(2*4.80*Sin53)
x = 0.056 m

Since x is positive, so spring is stretched

Part F:
Energy stored in spring, E = 0.5*k*x^2
E = 0.5 * 4.80 * 0.056^2
E = 0.0075264 J

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