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  • (a) State Biot – Savart law and express it in the vector form. (b) Using Biot – Savart law, obtain the expression for the magnetic field due to a circular coil of radius <img alt="r" src="https://entrancecorner.codecogs.com/g

(a) State Biot – Savart law and express it in the vector form.

(b) Using Biot – Savart law, obtain the expression for the magnetic field due to a circular coil of radius r, carrying a current I at a point on its axis distant x from the centre of the coil.

 

 
 
 
 
 

Answers (1)

(a) Biot- Savart law -

   The magnetic field due to a current element of length dl carrying a current I at a point P, which is at a distance r from it is 

(1) directly proportional to the current I and element length \vec{dl}.

(2) Inversely proportional to the square of the distance between the current element and the point P.

The direction of the magnetic field is perpendicular to the plane containing \vec{dl} and \vec{r}

Vector form -

 \vec{dB}=\frac{\mu _{0}}{4\pi }I\frac{\vec{dl}\times \vec{r}}{r^{3}}

(b)

Due to the current element dl considered at A the magnetic field at P is-

        dB=\frac{\mu _{0}I}{4\pi }\frac{\left | \vec{dl}\times \vec{r} \right |}{r^{3}}

        Since, dl\times r=dlr

         dB=\frac{\mu _{0}I}{4\pi }\frac{{dl} }{r^{2}}

        dB=\frac{\mu _{0}I}{4\pi }\frac{dl}{(X^{2}+R^{2})}

Consider a current element opposite to A, that is at B,then we can see that the y components of the magnetic, field due to this current element cancel and X component is only present.

So we can say that the net magnetic field is along the X-direction.

        dB_{X}=dB\cos \theta

     Net magnetic field at P is

        B=\int dB_{X}=\int dB\cos \theta

        =\int \frac{\mu _{0}}{4\pi }\frac{Idl}{(X^{2}+R^{2})}\cos \theta

        =\int \frac{\mu _{0}}{4\pi }\frac{Idl}{(X^{2}+R^{2})}\frac{R}{(X^{2}+R^{2})^{\frac{1}{2}}}

        =\frac{\mu _{0}I\; R}{4\pi \; (X^{2}+R^{2})^{\frac{3}{2}}}\int dl=\frac{\mu _{0}IR\; 2\pi R}{4\pi (X^{2}+R^{2})^{\frac{3}{2}}}

    B=\frac{\mu _{0}IR^{2}}{2(X^{2}+R^{2})^{\frac{3}{2}}}

Posted by

Safeer PP

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