Answer:
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, [tex]T_{max}=2.0\ Nm[/tex]
First, let us calculate the area of the coil.
Area is given by the formula:
[tex]A=\frac{\pi d^2}{4}\\\\A=\frac{3.14\times (0.062)^2}{4}\\\\A=0.00302\ m^2[/tex]
Now, the torque acting on the circular coil is given by the formula:
[tex]T=NIAB\sin\theta\\Where,\\B\to magnetic\ strength[/tex]
Now, torque is maximum when the angle [tex]\theta=90[/tex]° or [tex]\sin\theta=1[/tex]
Therefore, the maximum torque is given as:
[tex]T_{max}=NIAB\\\\B=\frac{T_{max}}{NIA}[/tex]
Now, plug in the given values and solve for 'B'. This gives,
[tex]B=\frac{2.0\ Nm}{250\times 3.9\ A\times 0.00302\ m^2}\\\\B=0.679\ T[/tex]
Therefore, the magnetic field strength is 0.679 T.
