Magnetic Force on Current Carrying Wires

FORCE ON A CURRENT CARRYING WIRE

We know that $\vec{F}=q\left(\vec{v}\times \vec{B}\right),$

We can say $d\vec{F}=dq\left[\vec{v}\times \vec{B}\right]$

$ord\vec{F}=dq\left[\frac{d\vec{\ell }}{dt}\times \vec{B}\right]=\frac{dq}{dt}\left[d\ell \times \vec{B}\right]$ $\Rightarrow d\vec{F}=id\vec{\ell }\times \vec{B}$

Actually, $d\vec{F}=i(d\vec{\ell }\times \vec{B})$gives the force on the charge carriers within the length $d\vec{\ell }$. However, this force is converted, by collisions, into a force on the wire as a whole, a force which, moreover, is capable of doing work on the wire. The net force on a wire is found by integrating along length. A corollary of this is there is no net force on a current carrying loop in a uniform magnetic field.

FLEMING'S LEFT-HAND RULE

The direction of the force $\vec{F}=i(\vec{L}\times B)$is given by the Fleming's left hand rule. Close your left fist and then, "shoot your index finger in the direction of the magnetic field. Relax your middle finger in the direction of the current. The force on the conductor is shown by the direction of the erect thumb.

Illustration 1: A wire bearing a current of 10A lies perpendicular to a uniform magnetic field. A force of 0.2 N is found to exist on a section of the wire 80cm long. Determine the magnetic induction B.

Solution: For a straight segment of wire of length L,

F = i$\ell$B sin 90° or 0.2 = 10(0.80)B B = 0.025 T

The direction of B will be normal to the plane of the force and the wire.

Illustration 2: At the equator, the earth's magnetic field is nearly horizontal, direction from the southern to northern hemisphere. Its magnitude is about 0.50 G. Find the force (direction and magnitude) on a 20–m wire carrying a current of 30A parallel to the earth (a) from east to west, (b) from north to south.

Solution: 1 Tesle = 10⁴ Gauss

(a) F = iLB sin q = 30(20)(5 × 10^–5)(1) = 0.030 N, down

(b) F = 30(20)(5 × 10^­–5)(0) = zero

FORCE BETWEEN TWO PARALLEL CURRENT CARRYING WIRES

To calculate the force between two infinite parallel current carrying wires separated by a distance r, we take an arbitrary pt. 'P' on the second wire; then the magnetic field at this point due to other wire is

$\vec{B}=\frac{{\mu }_0{i}_1}{2\pi r}(-\hat{k})$

Force on an elementary length of the second wire is

$d\vec{F}=i_{2}d\ell \vec{j}\times \frac{{\mu }_0{i}_1}{2\pi r}(-\hat{k})$ $\frac{d\vec{F}}{d\ell }=\frac{{\mu }_0{i}_1{i}_2}{2\pi r}(-\hat{i})$

We note that wires carrying current in the same direction attract each other.

(Verify using Fleming's left hand rule).

Illustration 3 : A long horizontal wire P carries a current of 50 A. It is rigidly fixed. Another fine wire Q is placed directly above and parallel to P. The weight of wire Q is 0.075 N m^–1 and carries a current of 25 A. Find the position of wire Q from P so that the wire Q remains suspended due to the magnetic repulsion. Also indicate the direction of current in Q with respect to P.

Solution: As force per unit length between two parallel current–carrying wires separated by a distance d is given by:

$\frac{dF}{dL}=\frac{{\mu }_0}{4\pi }\frac{2i_1{i}_2}{d}$

and is repulsive if the current in the wires is in opposite direction (otherwise attractive).

So, in order that wire Q may remain suspended, the force F on it must be repulsive and equal to its weight, i.e,. the current in the two wires must be in opposite directions and

F = mg, i.e., $\frac{F}{L}=\frac{Mg}{L}$

or, $\frac{{\mu }_0}{4\pi }\frac{2i_1{i}_2}{d}=\frac{Mg}{L}$ $\left[as\frac{dF}{dL}=\frac{{\mu }_0}{4\pi }\frac{2i_1{i}_2}{d}\right]$

or, $d={10}^{-7}\times \frac{2\times 50\times 25}{0.075}=\frac{1}{3}\times {10}^{-2}m$ $\left[as\frac{Mg}{L}=0.075\frac{N}{m}\right]$