Question

A coil of wire 8.6 cm in diameter has 15 turns and carries a current of 2.7 A. The coil is placed in a magnetic field of 0.56 T. What is the magnitude of the maximum torque that can be applied to the coil by the magnetic field?

Answer 1

Step-by-step explanation:

it is given that diameter = 8.6 cm

`radius =(8.6)/(2)=4.3\ cm=4.3* 10^(-2)\ m`

current =2.7 ampere

number of turns = 15

`area =\pi r^2=3.14* \left ( 4.3* 10^(-2) \right )^(2)=0.005806 m^(2)`

magnetic field =0.56 T

maximum torque= BINASINΘ for maximum torque sinΘ=1

so maximum torque==0.56×2.7×0.005806×15=0.13174 Nm