Solveeit Logo

Question

Question: The magnetic field on the axis of a current carrying circular coil of radius \[a\] at a distance \[2...

The magnetic field on the axis of a current carrying circular coil of radius aa at a distance 2a2a from its center will be
A. μ0i2\dfrac{{{\mu _0}i}}{2}
B. μ0i105a\dfrac{{{\mu _0}i}}{{10\sqrt 5 a}}
C. μ0i2a\dfrac{{{\mu _0}i}}{{2a}}
D. μ0i{\mu _0}i

Explanation

Solution

Use the expression for the magnetic field BB along the axis of the circular coil carrying an electric current at a distance d from the centre of the coil. This formula gives the relation between the electric current flowing through the circular coil, radius of the circular coil and the distance d from the centre of the circular coil. Substitute the value of the distance in this equation and derive the final required equation for the magnetic field.

Formula used:
The expression for the magnetic field BB along the axis of the circular coil carrying an electric current at a distance dd from the centre of the coil is given by
B=μ0iR22(d2+R2)32B = \dfrac{{{\mu _0}i{R^2}}}{{2{{\left( {{d^2} + {R^2}} \right)}^{\dfrac{3}{2}}}}} …… (1)
Here, μ0{\mu _0} is the permeability of the free space, ii is the electric current flowing through the circular coil, RR is the radius of the circular coil and dd is the distance of the point at which the magnetic field is to be measured from the centre of the coil.

Complete step by step answer:
We have given that the radius of a circular current carrying coil is aa.

Let the electric current flowing though this coil is ii.

Rewrite equation (1) for the magnetic field BB along the axis of the circular coil carrying an electric current.
B=μ0ia22(d2+a2)32B = \dfrac{{{\mu _0}i{a^2}}}{{2{{\left( {{d^2} + {a^2}} \right)}^{\dfrac{3}{2}}}}}

We have asked to determine the magnetic field at a distance 2a2a from the centre of the circular current carrying coil.

Substitute 2a2a for dd in the above equation.
B=μ0ia22((2a)2+a2)32B = \dfrac{{{\mu _0}i{a^2}}}{{2{{\left( {{{\left( {2a} \right)}^2} + {a^2}} \right)}^{\dfrac{3}{2}}}}}
B=μ0ia22(4a2+a2)32\Rightarrow B = \dfrac{{{\mu _0}i{a^2}}}{{2{{\left( {4{a^2} + {a^2}} \right)}^{\dfrac{3}{2}}}}}
B=μ0ia22(5a2)32\Rightarrow B = \dfrac{{{\mu _0}i{a^2}}}{{2{{\left( {5{a^2}} \right)}^{\dfrac{3}{2}}}}}
B=μ0ia22(5)32(a2)32\Rightarrow B = \dfrac{{{\mu _0}i{a^2}}}{{2{{\left( 5 \right)}^{\dfrac{3}{2}}}{{\left( {{a^2}} \right)}^{\dfrac{3}{2}}}}}
B=μ0ia22(5)32a3\Rightarrow B = \dfrac{{{\mu _0}i{a^2}}}{{2{{\left( 5 \right)}^{\dfrac{3}{2}}}{a^3}}}
B=μ0i2(55)a\Rightarrow B = \dfrac{{{\mu _0}i}}{{2\left( {5\sqrt 5 } \right)a}}
B=μ0i105a\therefore B = \dfrac{{{\mu _0}i}}{{10\sqrt 5 a}}
Therefore, the magnetic field BB along the axis of the circular coil carrying an electric current at a distance 2a2a from the centre of the circular coil is μ0i105a\dfrac{{{\mu _0}i}}{{10\sqrt 5 a}}.

Hence, the correct option is B.

Note: The students should keep in mind that the direction of the magnetic field asked in the above equation is given by the right hand thumb rule which stated that if the curled fingers give the direction of the electric field then the direction of the thumb represent the direction of the magnetic field. The same rule can be stated with a curled finger giving the direction of the magnetic field and the thumb representing the direction of the electric current.