Magnetics 6 Question 28
31. A current of $10 A$ flows around a closed path in a circuit which is in the horizontal plane as shown in the figure. The circuit consists of eight alternating arcs of radii $r _1=0.08 m$ and $r _2=0.12 m$. Each subtends the same angle at the centre.
(2001, 10M)
(a) Find the magnetic field produced by this circuit at the centre.
(b) An infinitely long straight wire carrying a current of $10 A$ is passing through the centre of the above circuit vertically with the direction of the current being into the plane of the circuit. What is the force acting on the wire at the centre due to the current in the circuit? What is the force acting on the $\operatorname{arc} A C$ and the straight segment $C D$ due to the current at the centre?
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Answer:
Correct Answer: 31. (a) $6.54 \times 10^{-5} T$ (vertically upward or outward normal to the paper) (b) Zero, Zero, $8.1 \times 10^{-6} N$ (inwards)
Solution:
- Given, $i=10 A, r _1=0.08 m$ and $r _2=0.12 m$. Straight portions i.e. $C D$ etc, will produce zero magnetic field at the centre. Rest eight arcs will produce the magnetic field at the centre in the same direction i.e. perpendicular to the paper outwards or vertically upwards and its magnitude is
$$ \begin{aligned} B & =B _{\text {inner arcs }}+B _{\text {outer arcs }} \\ & =\frac{1}{2} \frac{\mu _0 i}{2 r _1}+\frac{1}{2} \frac{\mu _0 i}{2 r _2}=\frac{\mu _0}{4 \pi}(\pi i) \frac{r _1+r _2}{r _1 r _2} \end{aligned} $$
Substituting the values, we have
$B=\frac{\left(10^{-7}\right)(3.14)(10)(0.08+0.12)}{(0.08 \times 0.12)} \Rightarrow B=6.54 \times 10^{-5} T$
(vertically upward or outward normal to the paper)
(b) Force on $A C$
Force on circular portions of the circuit i.e. $A C$ etc, due to the wire at the centre will be zero because magnetic field due to the central wire at these arcs will be tangential $\left(\theta=180^{\circ}\right)$.
Force on $C D$
Current in central wire is also $i=10 A$. Magnetic field at distance $x$ due to central wire
$$ B=\frac{\mu _0}{2 \pi} \cdot \frac{i}{x} $$
$\therefore$ Magnetic force on element $d x$ due to this magnetic field
$$ \begin{aligned} d F & =(i) \frac{\mu _0}{2 \pi} \cdot \frac{i}{x} \cdot d x=\frac{\mu _0}{2 \pi} i^{2} \frac{d x}{x} \\ (F & \left.=i l B \sin 90^{\circ}\right) \end{aligned} $$
Therefore, net force on $C D$ is
$$ F=\int _{x=r _1}^{x=r _2} d F=\frac{\mu _0 i^{2}}{2 \pi} \int _{0.08}^{0.12} \frac{d x}{x}=\frac{\mu _0}{2 \pi} i^{2} \ln \frac{3}{2} $$
Substituting the values
$$ \begin{aligned} F & =\left(2 \times 10^{-7}\right)(10)^{2} \ln (1.5) \\ \text { or } \quad F & =8.1 \times 10^{-6} N(\text { inwards }) \end{aligned} $$
Force on wire at the centre
Net magnetic field at the centre due to the circuit is in vertical direction and current in the wire in centre is also in vertical direction. Therefore, net force on the wire at the centre will be zero. $\left(\theta=180^{\circ}\right)$. Hence,
(i) force acting on the wire at the centre is zero.
(ii) force on $\operatorname{arc} A C=0$.
(iii) force on segment $C D$ is $8.1 \times 10^{-6} N$ (inwards).
