Question

A wire is bent into a circular coil of radius r =4.4 cm with 21 turns clockwise, then continues and is bentinto a square coil (length 2r) with 33 turns counterclockwise. A current of10.8 mA is running through the coil,and a 0.350 T magnetic field is applied to the plane of the coil.(a) What is the magnitude of the magnetic dipole moment of thecoil?μcoil = 3Your answer is incorrect. A ·m2Find the magnetic moments of the circular part and the square partseparately and combine them together to get the total magneticmoment. Your response differs from the correct answer by orders ofmagnitude.(b) What is the magnitude of the torque acting on the coil?

Answer 1

Answer:

a) The total magnetic dipole moment is 1.381x10⁻³A m²

b) The total torque dipole moment is 4.833x10⁻⁴N m

Explanation:

Given data:

r = radius = 4.4 cm = 0.044 m

N₁ = circular turns = 21

N₂ = square turns = 33

I = current = 10.8 mA = 10.8x10⁻³A

B = magnetic field = 0.35 T

a) The magnetic dipole moment (circular) is equal to:

$\mu _{1} =N_{1} LA=N_{1}I\pi r^{2} =21*10.8x10^{-3} *\pi *(0.044)^{2} =1.379x10^{-3} Am^{2}$

The magnetic dipole moment (square) is equal to:

$\mu _{2} =N_{2} LA=N_{2}I\pi L^{2} =33*10.8x10^{-3} *(0.088)^{2} =2.76x10^{-3} Am^{2}$

The total magnetic dipole moment is equal to:

$\mu _{total} =\mu _{2} -\mu _{1}=2.76x10^{-3} -1.379x10^{-3} =1.381x10^{-3}Am^{2}$

b) The torque dipole moment (circular) is equal to:

$\tau _{1} =\mu _{1} B=\mu _{1} Bsin90=1.379x10^{-3} *0.35*1=4.827x10^{-4} Nm$

The torque dipole moment (square) is equal to:

$\tau _{2} =\mu _{2} B=\mu _{2} Bsin90=2.76x10^{-3} *0.35*1=9.66x10^{-4} Nm$

The total torque dipole moment is equal to:

$\tau _{total} =\tau _{2} -\tau _{1}=9.66x10^{-4} -4.827x10^{-4} =4.833x10^{-4}Nm^{2}$