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

Answer:

Explanation:

it is given that diameter = 8.6 cm

$radius =\frac{8.6}{2}=4.3\ cm=4.3\times 10^{-2}\ m$

current =2.7 ampere

number of turns = 15

$area =\pi r^2=3.14\times \left ( 4.3\times 10^{-2} \right )^{2}=0.005806 m^{2}$

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