# A circular loop of radius a , carrying a current I ,is placed in a 2D magnetic field. The centre of the loop coincides with the centre of the field. The strength of magnetic field at the periphery of loop is B. Find the magnetic force on the wire?

Aug 24, 2016

This solution does not appear to be correct as the magnetic field assumed is not 2D but a 3D.
It may be ignored.

#### Explanation:

We know that Magnetic Force on a Current-Carrying Wire is perpendicular to both the wire and the magnetic field. The direction of magnetic force is given by the right hand rule or by cross product of the two vectors. It comes from the expression for ${\vec{F}}_{B}$ experienced by a charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$
${\vec{F}}_{B} = q \left(\vec{v} \times \vec{B}\right)$

As total Magnetic Force is the Sum of Forces on all charges in the wire, it can be shown for a current $I$ flowing in a wire of length $\vec{l} ,$ where length vector is directed along the direction of the electric current, that
${\vec{F}}_{B} = I \left(\vec{l} \times \vec{B}\right)$ ......(1)

For a circular loop of radius $a$ carrying current $I$.

Lets consider infinitesimal length $\mathrm{dv} e c s$ of the loop in the magnetic field $\vec{B}$. Magnetic Force on this length is
$\mathrm{dv} e c {F}_{B} = I \left(\mathrm{dv} e c s \times \vec{B}\right)$ .....(2)
It is given that the centre of the loop coincides with the centre of the 2D magnetic field. The strength of magnetic field at the periphery of loop is $| \vec{B} |$ as in the figure. We can therefore assume that the loop of wire is located in a uniform magnetic field $\vec{B}$

To find the total magnetic force we need to find the integral of equation (2) over the loop. We get
${\vec{F}}_{B} = I \left(\oint \text{ } \mathrm{dv} e c s\right) \times \vec{B}$ .....(3)
We see that infinitesimal vector lengths $\mathrm{dv} e c s$ form a closed polygon. Their vector sum $= 0$.
$\implies \oint \text{ } \mathrm{dv} e c s = 0$
Equation (3) becomes
${\vec{F}}_{B} = 0$
.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-

Hint

For each infinitesimal length of the current carrying loop, we can locate a diametrically opposite infinitesimal length carrying current in the opposite direction.

Aug 31, 2016  A circular loop of radius a, carrying a current I,is placed in a 2D magnetic field $\vec{B}$. The centre of the loop C coincides with the centre of the field. The strength of magnetic field at the periphery of loop is B. We are to find the magnetic force on the wire.

Let us consider an infinitesimal length $\mathrm{dv} e c s$ of the loop in the planer magnetic field $\vec{B}$. Magnetic Force on this length is

$\mathrm{dv} e c {F}_{B} = I \left(\mathrm{dv} e c s \times \vec{B}\right)$

As the angle between $\mathrm{dv} e c s \mathmr{and} \vec{B} \text{ is } \frac{\pi}{2}$

$\mathrm{dv} e c {F}_{B} = I \cdot \left\mid \mathrm{dv} e c s \right\mid \cdot \left\mid \vec{B} \right\mid \cdot \sin \left(\frac{\pi}{2}\right) \cdot \hat{e}$, where $\text{ "hate" }$ represents unit vector of force perpendicular to the plane of the loop.

So putting abs(dvecs)=ds , " " abs(vecB)=B and sin)pi/2=1 we get

$\mathrm{dv} e c {F}_{B} = I \cdot \left(\mathrm{ds}\right) \cdot B \cdot \hat{e} \ldots . . \left(1\right)$

By Fleming's Left Hand Rule or by Rule of cross product of two vector we can easily determine the direction of force experienced by the considered small length segment (ds) of the loop. This force will be directed perpendicular to the plane of the loop as shown in the figure.

If we are to find the total magnetic force on the loop , we are to add up the all magnetic forces acting on the all infinitesimal length segments constituting the loop.
It can be done by integrating equation (1) all over the circumference($2 \pi a$)of the loop

${\vec{F}}_{B} = = {\int}_{0}^{2 \pi a} I \cdot \left(\mathrm{ds}\right) \cdot B \cdot \hat{e}$

$\implies {\vec{F}}_{B} = B I {\int}_{0}^{2 \pi a} \left(\mathrm{ds}\right) \hat{e} = 2 \pi a B I \hat{e}$