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  • (a) Derive the expression for the magnetic field due to a current carrying coil of radius r at a distance x from the centre along the X-axis. (b) A straight wire carrying a current of 5 A is bent into a semicircular arc of radius 2 cm as show

(a) Derive the expression for the magnetic field due to a current carrying coil of radius r at a distance x from the centre along the X-axis.

(b) A straight wire carrying a current of 5 A is bent into a semicircular arc of radius 2 cm as shown in the figure. Find the magnitude and direction of the magnetic field at the centre of the arc.

 

 

 

 
 
 
 
 

Answers (1)

(a )

Consider a circular loop of radius R carrying current i. We have to find the magnetic field at a point P on the axis of the loop at a distance x from the centre.  Now, consider a current element id\vec l of the wire. The magnetic field at P due to this current element is :
      d\vec B=\frac{\mu_0}{4\pi}i\frac{d\vec l\times\vec r}{r^3}
d B=\frac{\mu_0}{4\pi}\frac{id l}{r^2}=\frac{\mu_0}{4\pi}\frac{id l}{\left ( R^2+X^2 \right )}  
The component along the x-axis is
dB\cos\theta=\frac{\mu_0 i}{4\pi}\frac{Rdl}{\left ( R^2+X^2 \right )^\frac{3}{2}}    
The two field due to the opposite elements have a resultant along the axis of the loop. Dividing the loop into such a pair of diametrically opposite elements, the resultant field at P must be along the x axis  and vertical components get cancelled
So, B_{total}=\int dB\cos\theta=\int \frac{\mu_0 iR}{4\pi}\frac{dl}{\left ( R^2+X^2 \right )^\frac{3}{2}} 
After integration,
B_{total}=\frac{\mu_0iR^2}{2\left ( R^2+X^2 \right )^{\frac{3}{2}}}
 

(b)Due to straight wire portion magneticfield=0 and the magnetic field is only due to the semicircular arc. 

B=\frac{\mu _{0}I}{4r}

      =\frac{4\pi \times 10^{-7}\times 5}{4\times 2\times10^{-2} }

        =7.85\times 10^{-5}T

The field is directed inwards perpendicular to the plane of the page.

Posted by

Safeer PP

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