First slide

Magnetic field due to a current carrying circular ring

Question

A closely wound flat circular coil of 25 turns of wire has diameter of 10 cm which carries current of 4 A. The flux density at the centre of a coil will be

Easy

A

$1 .679 \times 10^{- 5} T$
B

$1.256 \times 10^{- 3} T$
C

$1.512 \times 10^{- 5} T$
D

$2.28 \times 10^{- 4} T$

Solution

The magnetic field at the centre of the coil,
$B = \frac{μ_{0}}{4 π} \frac{2 πNi}{a}$
where N is number of turns, i is current and a is radius of coil.
$Given, N = 25, D = 10 cm = 10 \times 10^{- 2} m$
Radius of the coil,
$a = \frac{D}{2} = \frac{10 \times 10^{- 2}}{2} m = 0 .05 m$
And current i = 4 A. Putting these values in the expression for B , we get B=1.256 mT

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