A circular loop of wire having a radius of 5.58 cm carries a current of 0.230 A. A vector of unit length and parallel to the dipole moment of the loop is given by 0.60 – 0.80. If the loop is located in a uniform magnetic field given by = (0.140 T) + (0.109 T), find (a) the x-component, (b) the y-component, and (c) the z-component of the torque on the loop and (d) the magnetic potential energy of the

Respuesta :

Answer:

see explanation

Explanation:

Given quantities:

radius = r = 0.0558 [m]

current = I = 0.23 [A]

[tex]\vec{B} = <0.14[T] \hat i + 0.109[T] \hat j >[/tex]

Now we solve this by obtaining the torque acting on the dipole

[tex]\tau = M \times B[/tex]

We obtain the magnetic moment vector M, first, |M| is defined as [tex]M = IA[/tex], where A is the cross-section area of the loop which is [tex]A = \pi r^2=\pi (0.0558)^2= 0.00978 [m^2][/tex] then

[tex]|M| = 0.23*0.00978 = 0.00225 [A/m^2][/tex]

now the magnetic moment vector is equal to the magnetic dipole moment vector multiplied the magnitude we just obtained  [tex]\vec{M} = M \hat M[/tex]

[tex]= 0.00225 *<0.6 \hat i - 0.8 \hat j>[/tex]

Now:

a ) [tex]\tau = M \times B[/tex]

b) [tex]U = -M \cdot B[/tex]

a) the determinant gives us:

[tex]<0 \hat i,0 \hat j,0.00039915 \hat k>[/tex]

b) the dot product gives = [tex] -1*-7.2*10^{-6} = 7.2*10^{-6}[J][/tex]