Question

A wire carrying a steady current is first bent in form of a circular coil of one turn and then in form of a circular coil of two turns. Fine the ratio of magnetic fields at the centers of the two coils.

Answer 1

Answer:

1/4

Explanation:

pls find the solution

B= nμ₀i /2πr

where n=no of turns.

Answer 2

Answer:

The ratio of magnetic fields at the centers of the two coils.

About magnetic fields:

A magnetic field is a visual representation of how the magnetic force is distributed around and inside of magnetic objects. The most of us are at least somewhat familiar with common magnetic objects and are aware that there may be forces at work between them. We are aware that magnets have two poles and that the attraction or repulsion between two magnets depends on how they are oriented (similar poles). We are aware that this occurs in a certain area that surrounds a magnet. This area is described by the magnetic field.

Explanation:

Let the length of wire be l

When the wire is converted to coil of one turn

$l=2\pi r$

$r=\frac{l}{2\pi }$

$B=\frac{\mu_{0} }{2\pi } \frac{i}{r_{1}} =\frac{\mu_{0} }{2\pi } \frac{i}{\frac{l}{2\pi } } =\frac{\mu_{0}i }{l}$

When the wine is converted to coil of two turns

$l=4\pi r_{2}$

$r_{2}=\frac{l}{4\pi }$

$B_2=\frac{n\mu_0}{2\pi } \frac{l}{r_2 } = \frac{n\mu_0i}{2\pi \frac{l}{4\pi } }=\frac{n^2 \mu_0 i}{l}=\frac{2\times2 \mu_0 i}{l} =\frac{4\pi _0 i}{l}$

$\frac{B_1}{B_2}=\frac{1}{4}$

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