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

Derive an expression for the force acting between two very long parallel current carrying conductors and hence define the Ampere.

Accepted Answer

Consider two parallel current-carrying 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 magnetic 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.

On putting the value of $(\frac{μ_{0}}{2 π})$ in (vii) we can write:

$\frac{F}{L} = (2 \times 10^{- 7}) \times \frac{I_{a} I_{b}}{d}$ .....(viii)

$\frac{F}{L} = 2 \times 10^{- 7} N / m$ for $I_{a} = I_{b} = 1 A$ and $d = 1 m$ , Hence:

The ampere is the value of that steady current which, when maintained in each of the two very long, straight, parallel conductors of negligible cross-section, and placed one meter apart in vacuum, would produce on each of these conductors a force equal to $(2 \times 10^{- 7}) newtons per meter$ of length.

Derive an expression for the force acting between two very long parallel current carrying conductors and hence define the Ampere.

Important Questions on Moving Charges and Magnetism

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