An electric motor contains a 250-turn circular coil 6.2cm in diameter. . . If it is to develop a maximum torque of 2.0N*m at a current of 3.9A , what should be the magnetic field strength?

The magnetic field strength is 0.679 T.

Explanation:

Given:

Number of turns (N) = 250

Current in the coil (I) = 3.9 A

Diameter of the coil (d) = 6.2 cm =0.062 m [1 cm = 0.01 m]

Maximum torque is,

First, let us calculate the area of the coil.

Area is given by the formula:

Now, the torque acting on the circular coil is given by the formula:

Now, torque is maximum when the angle ° or

Therefore, the maximum torque is given as:

Now, plug in the given values and solve for 'B'. This gives,

Therefore, the magnetic field strength is 0.679 T.

force is mass*acceleration.

we have a very small mass: 11.4*0.001 kg

if we want 0.035n of force, we need an acceleration:

0.035 = 11.4*0.001*a

a = 0.035/(11.4*0.001) m/s^2

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this