Magnetism and Torque

Consider a point on the ring with coordinates:

$\vec{r} = (R\cos{\theta},R\sin{\theta},0)$

The arc length element is:

$d\vec{r} = (-R \ d\theta \ \sin{\theta},R \ d\theta \ \cos{\theta},0)$

The magnetic force on the this elementary arc length is:

$d\vec{F} = i (d\vec{r} \times \vec{B})$

Now consider the axis. It has a unit vector:

$\hat{a} = \left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0\right)$

Now, a vector perpendicular to the axis directed from the axis to the elementary arc length is:

$\vec{p} = \vec{r} - \left(\vec{r} \cdot \hat{a}\right) \hat{a}$

The reader may try to derive the above expression by themselves. Finally, the elementary torque due to the elementary force about the axis is:

$d\vec{\tau} = \vec{p} \times d\vec{F}$

Plugging in all expressions and simplifying, one gets:

$d\vec{\tau} = \left(\left(-\frac{B_oR^2i(\cos(2 \theta) + 5\sin(2 \theta) - 5)}{4}\right) \ \hat{i} + \left(-\frac{B_oR^2i(\cos(2 \theta) + 5\sin(2 \theta) - 5)}{4}\right) \ \hat{j} + \left(\frac{5B_oR^2i\cos(2\theta)}{2}\right) \ \hat{k}\right)d\theta$

Integrating the above:

$\vec{\tau} = \left(\frac{5\pi B_o R^2i}{2}\right) \left(\hat{i} + \hat{j}\right)$

So one can see that the net torque is along the $\hat{a}$ direction only. The resultant torque is:

$\lvert \vec{\tau} \rvert = \frac{5\pi B_o R^2i}{\sqrt{2}}$

The angular acceleration about this axis is:

$\alpha = \frac{2\lvert \vec{\tau} \rvert}{mR^2}$ $\implies \alpha = \left(5\pi \sqrt{2}\right) \frac{B_o i}{m}$ $\implies \boxed{\beta = 5\pi \sqrt{2}}$

Let us consider a point on the ring with position vector $\vec r=r(\hat i \cos \theta-\hat j \sin \theta)$

The unit vector along the axis of rotation is $\hat u=\dfrac {1}{\sqrt 2 }(\hat i+\hat j)$

Force of magnetic field on an element of length $dl=rd\theta$ at that point is

$d\vec F=id\vec l\times \vec B$

Moment of this force on that element about the origin is

$d\vec N=\vec r\times d\vec F$

So, moment of the force about the given axis is

$dN_{axis}=\hat u. d\vec N$ .

Finally, $d\vec l=\vec {dα}\times \vec r$

After performing all calculations, we get

$dN_{axis}=\dfrac {1}{\sqrt 2 }(2+\sin^2 \theta+5\sin \theta\cos \theta)$

Integrating within the limits $0$ to $2π$ , we obtain

$N_{axis}=\dfrac {5π}{\sqrt 2 }ir^2B_0$

So the equation of motion of the ring is

$\dfrac 12 mr^2α=\dfrac {5π}{\sqrt 2 }ir^2B_0$

$\implies α=5\sqrt 2π\times \dfrac {iB_0}{m}$

Therefore $β=5\sqrt 2π\approx \boxed {22.214}$ .

This was a fun one. I have attached some heavily commented solution code below.

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import math

Num = 10**6   # spatial resolution parameter

R = 1.0    # radius
B0 = 1.0   # mag flux density multiplier
i = 1.0    # current
m = 1.0    # loop mass

Bx = 2.0*B0   # magnetic flux density components
By = -3.0*B0
Bz = 5.0*B0

dtheta = 2.0*math.pi/Num   # incremental theta

#########################################

ux = 1.0/math.sqrt(2.0)   # unit vector along rotation axis
uy = 1.0/math.sqrt(2.0)
uz = 0.0

I = 0.5*m*(R**2.0)  # loop moment of inertia

#########################################

Tx = 0.0    # initialize torque vector
Ty = 0.0    # torque calculated with respect to principal axes x,y,z
Tz = 0.0

theta = 0.0

while theta <= 2.0*math.pi:   # integrate over loop

    x = R*math.cos(theta)   # spatial coordinates on loop
    y = R*math.sin(theta)
    z = 0.0

    dx = -R*math.sin(theta) * dtheta  # incremental changes in spatial coord
    dy = R*math.cos(theta) * dtheta   # components of dL vector
    dz = 0.0

    dFx = i*(dy*Bz - dz*By)      # force on infinitesimal length of loop
    dFy = i*(-(dx*Bz - dz*Bx))   # dF = i * (dL cross B)
    dFz = i*(dx*By - dy*Bx)

    dTx = y*dFz - z*dFy         # torque on infinitesimal length of loop
    dTy = -(x*dFz - z*dFx)      # dT = (position vec) cross dF
    dTz = x*dFy - y*dFx

    Tx = Tx + dTx  # update torque
    Ty = Ty + dTy
    Tz = Tz + dTz

    theta = theta + dtheta

#########################################

Tproj = Tx*ux + Ty*uy + Tz*uz   # find projection of torque vector......
                                # onto rotation axis


alpha = Tproj/I   # rotational Newton's 2nd Law to find ang acc

print Num
print alpha

# Print results

#>>> 
#10000
#22.2144146908
#>>> ================================ RESTART ================================
#>>> 
#100000
#22.2145924061
#>>> ================================ RESTART ================================
#>>> 
#1000000
#22.2144324623
#>>>