Question

A circular coil has a 18.0 cm radius and consists of 25.0 closely wound turns of wire. An externally produced magnetic field of magnitude 3.00 mT is perpendicular to the coil. (a) If no current is in the coil, what magnetic flux links its turns?

Answer 1

Answer:

The magnetic flux links to its turns = $7.6 \times10^{-3}$ Wb.

Explanation:

Given :

Radius of circular coil = $18 \times 10^{-2}$ m

Number of turns = 25

Magnetic field = $3 \times10^{-3}$ T

Magnetic flux (Φ) is a measure of the magnetic field lines passes through a given area. The unit of magnetic flux is weber (Wb).

We know that,

⇒    Φ = $BA$

Where $B =$ ext. magnetic field, $A =$ area of loop or coil.

But here given in question, we have turns of wire so our above eq. modified as follows.

⇒   Φ = $NBA$

Where $N =$ no. of turns.

∴    Φ = $25 \times 3 \times 10^{-3} \pi (18 \times10^{-2} )^{2}$

     Φ = $7.6 \times 10^{-3} Wb$

Thus, the magnetic flux links to its turns = $7.6 \times 10^{-3} Wb$