Consider two thin identical conducting wires covered with very th… - Vidyadip

MCQ

Consider two thin identical conducting wires covered with very thin insulating material. One of the wires is bent into a loop and produces magnetic field $B_1,$ at its centre when a current $I$ passes through it.The second wire is bent into a coil with three identical loops adjacent to each other and produces magnetic field $B_2$ at the centre of the loops when current $I/3$ passes through it. The ratio $B_1 : B_2$ is

A
$1:1$
✓
$1:3$
C
$1:9$
D
$9:1$

✓

Answer

Correct option: B.

$1:3$

b
For loop $\mathrm{B}=\frac{\mu_{0} \mathrm{nI}}{2 \mathrm{a}}$

where, $a$ is the radius of loop.

Then, $B_{1}=\frac{\mu_{0} I}{2 a}$

Now, for coil $\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{4 \pi} \cdot \frac{2 \mathrm{nA}}{\mathrm{x}^{3}}$

at the centre $x=$ radius of loop

$\mathrm{B}_{2} =\frac{\mu_{0}}{4 \pi} \cdot \frac{2 \times 3 \times(1 / 3) \times \pi(\mathrm{a} / 3)^{2}}{(\mathrm{a} / 3)^{3}} $

$=\frac{\mu_{0} \cdot 3 \mathrm{I}}{2 \mathrm{a}} $

$\therefore \frac{B_{1}}{B_{2}}=\frac{\mu_{0} I / 2 a}{\mu_{0} \cdot 3I / 2 a}$

$B_{1}: B_{2}=1: 3$

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