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.

speed

ratio of distance traveled to time interval

net force

vector sum of forces on object

object

source of diverging light rays either luminous or illuminated

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explanation:

massxgravity x heght=weightxheight= pe

4900=490xh

h=10