Question

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?

Answer 1

The magnetic field strength is 0.679 T.

Step-by-step 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,
`T_(max)=2.0\ Nm`

First, let us calculate the area of the coil.

Area is given by the formula:

`A=(\pi d^2)/(4)\\\\A=(3.14* (0.062)^2)/(4)\\\\A=0.00302\ m^2`

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

`T=NIAB\sin\theta\\Where,\\B\to magnetic\ strength`

Now, torque is maximum when the angle
`\theta=90`° or
`\sin\theta=1`

Therefore, the maximum torque is given as:

`T_(max)=NIAB\\\\B=(T_(max))/(NIA)`

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

`B=(2.0\ Nm)/(250* 3.9\ A* 0.00302\ m^2)\\\\B=0.679\ T`

Therefore, the magnetic field strength is 0.679 T.