Q. 445.0( 1 Vote )

# A circular coil of 200 turns has a radius of 10 cm and carries a current of 2.0 A.

(a) Find the magnitude of the magnetic field vector B at the center of the coil.

(b) At what distance from the center along the axis of the coil will the field B drop to half its value at the center?

(3√4 = 1.5874)

Answer :

Given:

Number of turns(n) = 200

Radius of circular coil(r) = 10 cm = 0.1 m

Current carried by the coil(i) = 2 A

Formula used:

**(a) Magnetic field at the center of a circular coil of n turns(B) =**

**M=****,**

where

μ_{0} = magnetic permeability of vacuum = 4π x 10^{-7} T m A^{-1},

i = current carried by coil,

r = radius of coil

Hence, from the given information,

Magnetic field at the center of the given coil

= T = 2.51 x 10^{-3} T = 2.51 mT

**(b) The magnetic field at a distance point in the coil is given by B _{p},**

The magnetic field at the center of the coil is

Given: the magnetic field at the center is 1/2 of the initial

The magnetic field at the center of the coil is

On equation the magnetic field at the center and the magnetic field at the any distant point from the coil, we get

On equating the above equation, we get

(r^{2} + x^{2})^{3} =4r^{6}

We get

x^{2} + r^{2}= 4^{1/3}r^{2}

x^{2}=0.58 r^{2}

x^{2}=0.5× 100=58

x=± 7.66 cm

Magnetic field will drop to half of its value at the center if the distance of that point from the center of the coil along the axis of coil is equal to 7.66 cm.

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