Wire loops…
Wire loops in an external magnetic field
One wire loops that is having a current 0f 140 mA and a radius of 0.06 m. This loop is situated in a magnetic field $B=0.4i+0.35j$ T and the unit vector normal to surface is $n=0.40i-0.92k$.
Please find the magnetic dipole moment. Find the i and j components of the torque and the potential energy of loop situated in this magnetic field.
The magnetic moment of a wire loop is the product betwenn its surface and its current. Thus:
$m = I*S =0.14*pi*0.06^2 =1.583*10^-3 A*m^2$
If we write this as vectors (for m, i and j) we have
$m = 1.583*10^{-3}*(0.4*i -0.92*j) =6.33*10^{-4}*j -1.46*10^{-3}*j$ $(A*m^2)$
The vector product between the magnetic moment of the wire loop and the external magnetic field in which this loop is situated defines the Torque as a vector. We write this vactor product as a matrix, having the i, j, k vectors on the first row, the components of the magnetic moment on the second row and the components of the magnetic moment of the third row:
$T = m x B =\begin{vmatrix} i & j& k\\
6.33*10^{-4}& -1.46*10^{-3}& 0\\
0.40& 0& 0.35\end{vmatrix}=$
$= i*(-0.511*10^-3) +j*(-2.216*10^-4) +k*(0.584*10^-3)$
$Tx =-0.511*10^{-3} N*m$
$Ty =-2.216*10^{-4} N*m$
$Tz =+0.584*10^{-3} N*m$
Magnetic potential energy is also defined as
$U =-m*B =-(6.33*10^{-4}*i -1.46*10^{-3}*i) *(0.4*i +0.35*k) =-6.33*10^{-4}*0.4 =-2.532*10^{-4} J$
Question 2
Around a 0.006 m pencil there are wrapped 20 loops of wire. Through this wire goeas a current of 3 A, when the pencil is located in a 5 T magnetic field. Please find the value of the torque on this pencil.
The magnetic moment $m$ of the solenoid is defined as
$m = I*N*S =3*20*\pi*(3*10^{-3})^2=1696.5*10^{-6} A*m^2$
The direction of this magnetic moment vector is perpendicular to the wire loops plane.
The torque of the external magnetic field is
$T = m \times B = m*B*\sin(\alpha) =1696.5*10^{-6}*5*\sin(60) = 7345.9*10^{-6} N*m$
(where $\times$ is the vector product)
