Name: Motion in Combined Electric and Magnetic Fields
Uploaded: 2019-07-19
Duration: 6 min 38 s
Description: Have you ever thought about the motion of charged particle when it passes through the combination of the electric and magnetic field? Let's watch this video...

Question

Deduce an expression for the force on a current -carrying conductor placed in a magnetic field. Derive an expression for the force per unit length between two parallel current-carrying conductors.

Accepted Answer

We know that a moving charge experiences a force in a magnetic field. Also, electric current is the rate of flow of charge. Hence, a current carrying conductor experiences a force in a magnetic field. Consider a rod of a uniform cross-sectional area $A$ and length $L$ . Let the number of electrons per unit volume in the conductor be $n$ . Hence, the total number of electrons in the conductor be $n L A .$ If the average drift velocity of each electron is $v_{d}$ , then the force $(F)$ on these charge carriers in a magnetic field $B$ :

$\vec{F} = (n L A) e {\vec{v}}_{d} \times \vec{B}$ .....(i)

The current $(I)$ in the conductor:

$I = n e A v_{d}$ .......(ii)

Using (i) and (ii), we get the force on a current-carrying straight conductor of length $L$ :

$\vec{F} = \vec{I L} \times \vec{B}$ ......(iii)

Consider two parallel current-carriyng long conductors $a$ and $b$ separated by a distance $d$ . Let the current through them be $I_{a}$ and $I_{b}$ respectively. Let these wires lie in the plane of paper. The conductor $a$ produces the same magnetic field at all points on conductor $b$ as both the wires are parallel.

The direction of this field will be inside the plane of the paper. The magnitude of this magnetic field can be written using Ampere's circuital law:

$B_{a} = \frac{μ_{0} I_{a}}{2 π d}$ ......(iv)

Due to this field the conductor $b$ will experience a force towards $a$ , that has a magnitude:

$F_{b a} = I_{b} L B_{a}$ .......(v)

Using (iv) and (v), we get:

$F_{b a} = \frac{μ_{0} I_{a} I_{b}}{2 π d} L$ .....(vi)

The magnitude of force experienced by the current-carrying conductor $a$ due to $b$ is equal to given by (vi). But the direction will be the opposite.

Hence, the magnitude of the force per unit length between the two parallel current-carrying wires of length $(L)$ at a distance $(d)$ :

$\frac{F}{L} = \frac{μ_{0} I_{a} I_{b}}{2 π d}$ .....(vii)

where $I_{a}$ and $I_{b}$ are current flowing through them.

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