Magnetics Ques 69
- Three infinitely long thin wires, each carrying current $i$ in the same direction, are in the $x-y$ plane of a gravity free space. The central wire is along the $y$-axis while the other two are along $x= \pm d$.
$(1997,5 M)$
(a) Find the locus of the points for which the magnetic field $B$ is zero.
(b) If the central wire is displaced along the $z$-direction by a small amount and released, show that it will execute simple harmonic motion. If the linear density of the wires is $\lambda$, find the frequency of oscillation.
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Answer:
Correct Answer: 69.(a) $x=0=z$ and $z=0, x= \pm \frac{d}{\sqrt{3}}$ (b) $f=\frac{i}{2 \pi d} \sqrt{\frac{\mu _0}{\pi \lambda}}$
Solution:
Formula:
Magnetic Force Acting On A Current Carrying Wire:
- Magnetic field will be zero on the $y$-axis i.e.
Magnetic field cannot be zero in region I and region IV because in region I magnetic field will be along positive $z$-direction due to all the three wires, while in region IV magnetic field will be along negative $z$-axis due to all the three wires. It can zero only in region II and III.
Let magnetic field is zero on line $(z=0)$ and $x=x$. Then magnetic field on this line due to wires 1 and 2 will be along negative $z$-axis and due to wire 3 along positive $z$-axis. Thus
$$ \begin{aligned} B _1+B _2 & =B _3 \\ \text { or } \quad \frac{\mu _0}{2 \pi} \frac{i}{d+x}+\frac{\mu _0}{2 \pi} \frac{i}{x} & =\frac{\mu _0}{2 \pi} \frac{i}{d-x} \\ \text { or } \quad \frac{1}{d+x}+\frac{1}{x} & =\frac{1}{d-x} \end{aligned} $$
This equation givesz $x= \pm \frac{d}{\sqrt{3}}$
where magnetic field is zero.
(b) In this part, we change our coordinate axes system, just for better understanding.
There are three wires 1,2 and 3 as shown in figure. If we displace the wire 2 towards the $z$-axis, then force of attraction per unit length between wires ( 1 and 2$)$ and (2 and 3 ) will be given as
The components of $F$ along $x$-axis will be cancelled out. Net resultant force will be towards negative $z$-axis (or mean position) and will be given by
$$ \begin{aligned} & F _{\text {net }}=\frac{\mu _0}{2 \pi} \frac{i^{2}}{r}(2 \cos \theta)=2 \frac{\mu _0}{2 \pi} \frac{i^{2}}{r} \frac{z}{r} \\ & F _{\text {net }}=\frac{\mu _0}{\pi} \frac{i^{2}}{\left(z^{2}+d^{2}\right)} z \end{aligned} $$
If $z \ll d$, then
$$ z^{2}+d^{2}=d^{2} \text { and } F _{\text {net }}=-\frac{\mu _0}{\pi} \frac{i^{2}}{d^{2}} \cdot z $$
Negative sign implies that $F _{\text {net }}$ is restoring in nature Therefore, $F _{\text {net }} \propto-z$
i.e. the wire will oscillate simple harmonically.
Let $a$ be the acceleration of wire in this position and $\lambda$ is the mass per unit length of this wire then
$$ F _{\text {net }}=\lambda \cdot a=-\frac{\mu _0}{\pi} \frac{i^{2}}{d^{2}} \quad z \quad \text { or } a=-\frac{\mu _0 i^{2}}{\pi \lambda d^{2}} \cdot z $$
$\therefore$ Frequency of oscillation
$$ \begin{aligned} f & =\frac{1}{2 \pi} \sqrt{\frac{\text { acceleration }}{\text { displacement }}}=\frac{1}{2 \pi} \sqrt{\frac{a}{z}} \\ & =\frac{1}{2 \pi} \frac{i}{d} \sqrt{\frac{\mu _0}{\pi \lambda}} \text { or } \quad f=\frac{i}{2 \pi d} \sqrt{\frac{\mu _0}{\pi \lambda}} \end{aligned} $$
BETA
