Figure shows a wire ring of radius$\mathit{a}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{8}\mathbf{\text{cm}}$that is perpendicular to the general direction of a radially symmetric, diverging magnetic field. The magnetic field at the ring is everywhere of the same magnitude $\mathit{B}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{4}\mathbf{\text{mT}}$, and its direction at the ring everywhere makes an angle $\mathit{\theta }\mathbf{=}\mathbf{20}\mathbf{°}$with a normal to the plane of the ring. The twisted lead wires have no effect on the problem. Find the magnitude of the force the field exerts on the ring if the ring carries a current $\mathit{i}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{6}\mathbf{\text{mA}}$.

a) The radius of wire ring is $a=1.8\text{cm or}1.8\times {10}^{-2}\text{m}$

b) The magnitude of the magnetic field is $B=3.4\text{mT or}3.4\times {10}^{-3}\text{T}$

c) The direction of the magnetic field at the ring with the normal of the plane, $\theta =20°$

d) The current flowing through the ring, $i=4.6\text{mA or}4.6\times {10}^{-3}\text{A}$

If a particle is moving with a uniform velocity within a uniform magnetic field, then it experiences a magnetic force due to its charge value that induces a current within the loop. The force acting on the particle is due to the current along the length of the conductive wire. The direction of this magnetic force is given by Fleming's right-hand rule as the force is perpendicular to both the speed and magnetic field acting on it.

The magnetic force along a loop wire is as follows:

$\overrightarrow{d{F}_B}=i\left(\overrightarrow{dL}\times \overrightarrow{B}\right)$ …… (i)

Here, $i$is the current in the loop, $\overrightarrow{B}$is the magnetic field vector that it experiences, $\overrightarrow{dL}$ is the length vector of the conducting wire

Consider an infinitesimal segment of the loop of length$ds$.

The magnetic field is perpendicular to the segment. Thus, the expression of magnitude of magnetic force on the segment is given using equation (i) and our assumption as follows:

$dF=iBds$

The horizontal component of the force has magnitude that is given as:

$d{F}_h=\left(iB\cos \theta \right)ds$

It has direction towards the center of the loop.

The vertical component of the force has magnitude that is given as:

$d{F}_y=\left(iB\sin \theta \right)ds$

It has upward direction. The net force on the loop is the sum of the forces on all the segments.

The horizontal components of the forces from all the vectors are both towards the center of the loop and in the opposite direction, hence, cancelled. Thus, the horizontal force component will yield a value of$0\text{N}$.

Being in the one direction, all the vertical components of the forces is now added to given a total vertical force value that is given by integrating the vertical component as follows:

$$\begin{gathered} F=\int d{F}_y \\ =\left(iB\sin \theta \right)\int ds \\ =\left(iB\sin \theta \right)\left(2\pi a\right) \\ =2\pi aiB\sin \theta \end{gathered}$$

Substitute the values and solve as:

$$\begin{gathered} F=2\times 3.14\times 0.018\text{m}\times 4.6\times {10}^{-3}\text{A}\times 3.4\times {10}^{-3}\text{T}\left(\sin 20°\right) \\ =6.0\times {10}^{-7}\text{N} \end{gathered}$$

The net force is the only vertical force acting on the loop.

Hence, the magnitude of the force is $6.0\times {10}^{-7}\text{N}$.