Answer to Question #116000 in Electric Circuits for Opeyemi Owonibi

Question

Ampere’s law assignment
Two long parallel wires are 10 cm apart and carry currents of 6 A and 4 A. what is the field B at 4 A carrying wire due to 6 A wire.

Answer 1

Explanations & calculations

Consider the sketch attached.

• Consider of the magnetic field induced on the point P by AB finite length conductor.
• In this question Biot - Savart law is applied which is

$\qquad\qquad dB = \frac{\mu_0i}{4\pi}\bigg(\frac{ldl*\sin\theta}{r^2}\bigg) \cdots(1)$ ; symbols have the standard meanings

Proof:

• In the equation above the integral is ldl & have 3 variables within the brackets, which should be reduced to evaluate (lets turn it into an integral of d$\small \theta$ )
• $\qquad\qquad \begin{aligned} \small r &= \small \frac{d}{\sin\theta_1}\cdots(2)\\ \small l &= \small \frac{d}{\tan\theta_1} = d\cot\theta\cdots(3)\\ \end{aligned}$
• Now find dl with d$\small \theta$ by differentiating (3) w.r.t $\theta$

$\qquad\qquad\qquad \begin{aligned} \small dl &= \small -\frac{d}{\sin^2\theta}*d\theta \end{aligned}$

• Now re-writing equation (1) we get,

$\qquad\qquad \begin{aligned} \small dB &= \small \frac{\mu_0i}{4\pi}\bigg(\frac{(\frac{-d}{\sin^2\theta_1}*d\theta)\sin \theta }{(\frac{d}{\sin\theta_1})^2}\bigg)\\ &= \small \frac{\mu_0i}{4\pi}\bigg(\frac{-\sin\theta_1}{d}d\theta\bigg)\\ &= \small \frac{-\mu_0i}{4\pi d}\big(\sin\theta d\theta\big) \end{aligned}$

• Now integrate over the AB length of the conductor to obtain the net magnetic field on P

$\qquad\qquad \begin{aligned} \small \intop dB &= \small \frac{-\mu_0i}{4\pi d} \intop_{\theta_1}^{(\pi-\theta_2)} \sin\theta d\theta\\ \small B &= \small \frac{-\mu_0i}{4\pi d}\big(-\cos\theta_2 - \cos\theta_1\big)\\ &= \small \frac{\mu_0i}{4\pi d}\big(\cos\theta_2 + \cos\theta_1\big) \end{aligned}$

• Now this is for the effect from a finite length conductor & as the conductor is infinite, $\theta_1,\theta_2 \to0 \,\,\text{and}\,\, \cos\theta_1,\cos\theta_2\to 1$

• Therefore, magnetic field strength on a point due to a conductor of infinite length becomes,

$\qquad\qquad\qquad \begin{aligned} \small B = \small \frac{\mu_0i}{2\pi d} \end {aligned}$

Answer

• Magnetic field on 4A carrying conductor due to the other conductor,

$\qquad\qquad \begin{aligned} \small B &= \small \frac{4\pi\times10^{-7}TmA^{-1}\times6A}{2\pi\times0.1m}\\ &= \small \bold{1.2\times10^{-5}T} \end{aligned}$