Question

An infinitely long current carrying wire and a small current carrying loop are in the plane of the paper as shown. The radius of the loop is a and distance of its centre from the wire is (d>>a). If the loop applies a force F on the wire then:

• A. $F\propto {\left(\frac{a}{d}\right)}^2$
• B. $F\propto \left(\frac{a^2}{d^3}\right)$
• C. $F\propto \left(\frac{a}{d}\right)$
• D. F=0

Answer 1

The correct option is A $F\propto {\left(\frac{a}{d}\right)}^2$

$\text{Given:-}$Radius of the loop = a

distance of its centre from the wire = (d>>a)

$\text{To find:-}$ the relation of the force with a and d.

$\text{Solution:-}$

The radius of this loop is given as a and the distance of separation of the wire with the center of the loop given as d. We are also told that (d>>a)

Here the circular loop is acting as a magnetic dipole and we know that the force due to magnetic dipole is given by,

$F=m.\nabla B$

Where, m is the magnetic moment of the dipole and B is the magnetic field due to the current carrying wire at the point at which the loop is kept.

Since we are dealing with one dimension, we could rewrite the equation as,

$F=m.\frac{dB}{dx}$................ (1)

We know that magnetic field due to current carrying wire at a distance x is given by,

$B=\frac{{\mu }_{0}I}{2\pi x}$............... (2)

We know that magnetic moment m due the given loop will be,

$m=NIA$

$ButN=1$ $and$

$A=\pi a^2\Rightarrow m=I\pi a^2$................ (3)

Substituting (2) and (3) in (1), we get,

$F=\left(I\pi a^2\right)\frac{d\left(\frac{{\mu }_{0}I}{2\pi x}\right)}{dx}\Rightarrow F=\left(I\pi a^2\right)(-\frac{{\mu }_{0}I}{2\pi x^2})$

But x here is given as d, so,

$F=-\frac{{\mu }_0{I}^2{a}^2}{2d^2}$

Therefore, we found that the force applied by the loop F,

$F\propto \frac{a^2}{d^2}$

$\text{Hence the correct option is A}$