Question

A closely wound flat circular coil of 25 turns of
wire has diameter of 10 cm and carries a current
of 4 ampere. Determine the magnetic flux density
at the centre of the coil :-​

Answer 1

Answer:

10^-5 Weber is the answer...

Answer 2

Magnetic flux density, $\phi=3.94\times 10^{-7}\ Wb$

Explanation:

it is given that,

Number of turns of the coil, N = 25

Diameter of the wire, d = 10 cm

Radius, r = 5 cm = 0.05 m

Current, I = 4 A

The magnetic field at the center of the coil is given by :

$B=\dfrac{\mu_oI}{2r}$

$B=\dfrac{4\pi \times 10^{-7}\times 4}{2\times 0.05}$

$B=5.02\times 10^{-5}\ T$

Let $\phi$ is the magnetic flux density  at the center of the coil. It is given by :

$\phi=BA$

$\phi=\pi r^2 B$

$\phi=\pi (0.05)^2 \times 5.02\times 10^{-5}$

$\phi=3.94\times 10^{-7}\ Wb$

So, the magnetic flux density  at the center of the coil is $3.94\times 10^{-7}\ Wb$. Hence, this is the required solution.

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Magnetic flux density

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