Question

Find Magnetic Field at The Centre ?​

Answer 1

${\small{\bf{ \orange{ \bigstar Magnetic \ field\ due\ to \ circular\ loo }}}}$${\small{\bf{ \orange{ p \bigstar }}}}$

(refer the attachment)

Let the small current element dB which is perpendicular to the centre at O,

be the radius = a

we know that biot savart law's formula,

$\bf{ dB = \dfrac { \mu_o }{4\pi }. \dfrac{ I \times dl \times sin \theta }{ r^2 } }$

${\theta = 90^0 ,sin90^0 = 1}$

so we get,

$\bf{ dB = \dfrac { \mu_o }{4\pi }. \dfrac{ I \times dl }{ a^2 } }$

now we will find the $B_{net }$ for whole loop,

$\bf{B_{net} = \int \dfrac { \mu_o }{4\pi }. \dfrac{ I \times dl }{ a^2 } }$

$\bf{B_{net} = \dfrac{\mu _o }{4\pi } \dfrac{I}{a^2} \int dl }$

$\sf{we \ know \ dl \ = 2\pi r = 2\pi a}$

$\bf{B_{net} = \dfrac{\mu _o }{4\pi } \dfrac{I}{a^2} 2 \pi a }$

$\boxed{\pink{\bf { B_{net} = \dfrac{\mu_o}{2 } \dfrac{I}{a} }}}$

this is the expression for centre of the loop

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Extra info-

For Semi circular loop expression is

$\tt {B= \dfrac{\mu_o}{4} \dfrac {I}{a} }$

For quadratic circular loop expression is

$\tt {B= \dfrac{\mu_o}{8} \dfrac {I}{a} }$

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Answer 2

This attachment will help you...