Question

The magnetic field B at a large distance a on the axis of a current carrying circular coil of small radius depends as-
कम त्रिज्या की धारावाही वृत्तीय कुण्डली के अक्ष पर चुम्बकीय क्षेत्र B अत्यधिक दूरी a पर किसके अनुसार निर्भर करता है-
O (A) Bas
1
O (B) Box
as
O (C) Baal
1
O (D) Br
a2​

Answer 1

To find:

The magnetic field intensity on the axis of the circular coil ?

Calculation:

According to Biot-Savart's Law:

$\displaystyle \therefore \: B = \int \: dB$

$\displaystyle \implies \: B = \int \: \dfrac{ \mu_{0}}{4\pi} \times \frac{I\:dl \: \sin( \alpha ) }{ {r}^{2} }$

$\displaystyle \implies \: B = \int \: \dfrac{ \mu_{0}}{4\pi} \times \frac{I\:dl }{ {r}^{2} } \times \frac{R}{r}$

$\displaystyle \implies \: B = \int \: \dfrac{ \mu_{0}}{4\pi} \times \frac{IR \times dl }{ {r}^{3} }$

$\displaystyle \implies \: B = \dfrac{ \mu_{0}}{4\pi} \times \frac{IR }{ {r}^{3} } \times \int dl$

$\displaystyle \implies \: B = \dfrac{ \mu_{0}}{4\pi} \times \frac{IR }{ {r}^{3} } \times (2\pi R)$

$\displaystyle \implies \: B = \dfrac{ \mu_{0}}{4\pi} \times \frac{IR }{ {( \sqrt{ {x}^{2} + {R}^{2} }) }^{3} } \times (2\pi R)$

$\displaystyle \implies \: B = \dfrac{ \mu_{0}}{4\pi} \times \frac{2\pi I{R}^{2} }{ {( \sqrt{ {x}^{2} + {R}^{2} }) }^{3} }$

$\displaystyle \implies \: B = \dfrac{ \mu_{0}}{4\pi} \times \frac{2\pi I{R}^{2} }{ {({x}^{2} + {R}^{2} )}^{ \frac{3}{2} } }$

For x >> R :

$\displaystyle \implies \: B = \dfrac{ \mu_{0}}{4\pi} \times \frac{2\pi I{R}^{2} }{ {({x}^{2} )}^{ \frac{3}{2} } }$

$\displaystyle \implies \: B = \dfrac{ \mu_{0}}{4\pi} \times \frac{2\pi I{R}^{2} }{ {x}^{3} }$

So, final answer is:

$\boxed{ \bf \: B = \dfrac{ \mu_{0}}{4\pi} \times \frac{2\pi I{R}^{2} }{ {x}^{3} } }$