Can someone please help with this problem

figure:

http://session.masteringphysics.com/problemAsset/1…

The 20.0cm by 35.0cm rectangular circuit shown in the figure, is hinged along side . It carries a clockwise I = 5.05A current and is located in a uniform 1.30T magnetic field oriented perpendicular to two of its sides, as shown in the figure.

a) Calculate only the force that exert the torque mentioned in part A in (N).

b) Use your results from part B to calculate the torque that the magnetic field exerts on the circuit about the hinge axis ab in (Nm).

### 4 Answers

a) to calculate the force on any side of the coil use the formula Force = magnetic field strength x current x length of side x sin(theta). The value of theta in this case is 90 as it is perpendicular to the magnetic field. The length of the side must be in metres. so the force on each side would be F = 1.3 x 5.05 x 0.2 x sin90 = 1.313N. Note i used the side of length 20cm because thats the side that will rotate due to the direction of the magnetic field.

b) The torque is calculated using torque = magnetic field x current x area of coil x cos(theta). the area must be in metres^2. T = 1.3 x 5.05 x (0.2×0.35) x cos0 = 0.46Nm. Use cos 0 because the angle between the hinge and the magnetic field before the coil rotates is 0.

hi! The solutions sound like a Boston lawyer? tension = something moved via capability, kinetic or otherwise. Torque = all the above on the tip of a 2nd arm. no longer something complicated and universal used 3000 + years in the past via lazy slaves? Then somebody rested the pole on a rock and lower back cool? A swing grew to become into discovered.

a) force will be exerted only on cd and ab segment since they both carry current in a direction perpendicular to that of the magnetic field. force on part cd = force on part ab = B*I*L = (1.3)*(5.05)*(0.2) = 1.313 N .. but by fleming’s rule, you can see that force on cd and ab are in opposite directions, so, net force on loop = 0

b) torque = force on cd * perpendicular distance from axis.

= 1.3*5.05*0.2*0.35 = 0.4955 Nm

Torque = force * radius