Induced Voltage Exercise (Part 3)

A general outline of the solution only provided. This was a fun problem, as always.

First, one must find the equation of the plane on which the loop lies. Consider a point $(x,y,z)$ on the plane, then it follows that any vector on the plane is normal to the given normal vector. This implies:

$\left((x-1)\hat{i} +(y-1)\hat{j}+(z-1)\hat{k}\right) \cdot \left(1\hat{i} +2\hat{j}+3\hat{k}\right) = 0$ $\implies x+2y+3z=6$

Consider any point that satisfies the above equation. I chose $(1,2,1/3)$ . A vector starting from the centre of the circle until that point is:

$\vec{n}_1 = \left(\hat{i} +2\hat{j}+\frac{1}{3}\hat{k}\right) - \left(\hat{i} +\hat{j}+\hat{k}\right)$ $\hat{n}_1 = \frac{3\left(\hat{j}-\frac{2}{3}\hat{k}\right)}{\sqrt{13}}$

Now, the given normal vector is:

$\vec{n} = \hat{i} +2\hat{j}+3\hat{k}$ $\hat{n}= \frac{\hat{i} + 2\hat{j}+3\hat{k}}{\sqrt{14}}=n_x \hat{i} + n_y \hat{j} + n_z \hat{k}$

Now, a unit vector perpendicular to both $\hat{n}_1$ and $\hat{n}$ is:

$\hat{n}_2 = \hat{n} \times \hat{n}_1$

Having found these directions, a point within the circular loop can be defined as follows:

$\vec{r} = x\hat{i} +y\hat{j}+z\hat{k}=\left(1\hat{i} +1\hat{j}+1\hat{k}\right) + r\cos{\theta} \ \hat{n}_1 + r\sin{\theta} \ \hat{n}_2$

An elementary area vector normal to this circular loop is:

$d\vec{S} = r \ dr \ d\theta \ \hat{n} = r \ dr \ d\theta\left(n_x \hat{i} + n_y \hat{j} + n_z \hat{k}\right)$

Now, using the definition of flux:

$d\Phi = \vec{B} \cdot d\vec{S}$

$d\Phi = \left(B_x n_x + B_y n_y + B_z n_z\right)r \ dr \ d\theta$

Substituting expressions and simplifying gives:

$d\Phi = t^2 \left(f_x(r,\theta)r \ dr \ d\theta\right) + t^3 \left(f_y(r,\theta)r \ dr \ d\theta\right) + \sin(2t) \left(f_z(r,\theta)r \ dr \ d\theta\right)$

$\Phi = t^2 \left(\int_{0}^{1} \int_{0}^{2 \pi}f_x(r,\theta)r \ d\theta \ dr\right) + t^3 \left(\int_{0}^{1} \int_{0}^{2 \pi}f_y(r,\theta)r \ d\theta \ dr\right)+\sin(2t) \left(\int_{0}^{1} \int_{0}^{2 \pi}f_z(r,\theta)r \ d\theta \ dr\right)$

Each of these integrals is evaluated numerically. This leads to:

$\Phi = At^2 + Bt^3+C\sin(2t)$ $\frac{d\Phi}{dt} = 2At+3Bt^2 + 2C\cos(2t)$

Substitute $t=2$ to get the required answer of $\approx 12.32$