In a certain region, a hypothetical magnetic field is described by...

The position vector of any point on the circular loop is given by:

$\vec{r} = a \cos\theta \hat{i} + a \sin\theta \hat{j} + d \hat{k}$

The distance of any point on the loop from the origin is:

$r = |\vec{r}| = \sqrt{a^2 + d^2}$

The unit vector in the radial direction is $\vec{e_r} = \dfrac{\vec{r}}{r}$. Thus, the magnetic field at the loop is:

$\vec{B} = B_0 \vec{e_r} = \dfrac{B_0}{\sqrt{a^2 + d^2}} (a \cos\theta \hat{i} + a \sin\theta \hat{j} + d \hat{k})$

The current element vector $d\vec{l}$ along the loop is:

$d\vec{l} = d\vec{r} = (-a \sin\theta \hat{i} + a \cos\theta \hat{j}) d\theta$

The magnetic force on this small current element is $d\vec{F} = i(d\vec{l} \times \vec{B})$.

Calculating the cross product:

$d\vec{l} \times \vec{B} = \dfrac{B_0 d\theta}{\sqrt{a^2 + d^2}} \begin{vmatrix} \hat{i} \text{ and } \hat{j} \text{ and } \hat{k} \\ -a \sin\theta \text{ and } a \cos\theta \text{ and } 0 \\ a \cos\theta \text{ and } a \sin\theta \text{ and } d \end{vmatrix}$

$d\vec{l} \times \vec{B} = \dfrac{B_0 d\theta}{\sqrt{a^2 + d^2}} [ (a d \cos\theta) \hat{i} - (-a d \sin\theta) \hat{j} + (-a^2 \sin^2\theta - a^2 \cos^2\theta) \hat{k} ]$

$d\vec{l} \times \vec{B} = \dfrac{B_0 d\theta}{\sqrt{a^2 + d^2}} (a d \cos\theta \hat{i} + a d \sin\theta \hat{j} - a^2 \hat{k})$

The total force on the loop is obtained by integrating $d\vec{F}$ from $\theta = 0$ to $\theta = 2\pi$:

$\vec{F} = \int_{0}^{2\pi} \dfrac{i B_0}{\sqrt{a^2 + d^2}} (a d \cos\theta \hat{i} + a d \sin\theta \hat{j} - a^2 \hat{k}) d\theta$

Since $\int_{0}^{2\pi} \cos\theta d\theta = 0$ and $\int_{0}^{2\pi} \sin\theta d\theta = 0$, the $\hat{i}$ and $\hat{j}$ components vanish.

$\vec{F} = \dfrac{i B_0}{\sqrt{a^2 + d^2}} (-a^2 \hat{k}) \int_{0}^{2\pi} d\theta = -\dfrac{2\pi a^2 i B_0}{\sqrt{a^2 + d^2}} \hat{k}$

The magnitude of the magnetic force exerted on the loop is:

$|\vec{F}| = \dfrac{2\pi a^2 i B_0}{\sqrt{a^2 + d^2}}$