Question

Class 12th Physics!
________

Derive an expression for the magnetic field due to straight current carrying conductor ?
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Answer 1

Answer:

plastic answer

Explanation:

plastic

Answer 2

EXPLANATION.

Expression for the magnetic field due to straight current carrying conductor.

As we know that,

⇒ θ + Φ + 90° = 180°.

⇒ θ = 90° - Ф.

BIOT savart law.

|dB| due to current element.

$\sf \implies dB = \dfrac{\mu_{o}}{4 \pi} \dfrac{idl sin \theta}{r^{2} }$

$\sf \implies dB = \dfrac{\mu_{o}}{4 \pi} \dfrac{i dl sin(90 - \phi )}{r^{2} }$

$\sf \implies dB = \dfrac{\mu_{o}}{4 \pi} \dfrac{i dl cos (\phi)}{r^{2} }$

Net magnetic field (B).

$\sf \implies B = \displaystyle \int dB = \int \dfrac{\mu_{o}}{4 \pi} \dfrac{idl cos (\phi)}{r^{2} }$

$\sf \implies B = \displaystyle \dfrac{\mu_{o}i}{4 \pi } \int \dfrac{dl cos (\phi)}{r^{2} }$

In ΔAPB.

⇒ cos(Ф) = a/r.

⇒ r = a/cos(Ф).

$\sf \implies B = \displaystyle \dfrac{\mu_{o} i}{4 \pi} \int \dfrac{dl cos (\phi)}{r^{2} }$

$\sf \implies B = \displaystyle \dfrac{\mu_{o} i}{4 \pi} \int \dfrac{dl cos(\phi) \times cos^{2} (\phi)}{a^{2} }$

$\sf \implies tan(\phi) = \dfrac{L}{a}$

$\sf \implies L = a \ tan(\phi)$

$\sf \implies \dfrac{dL}{d(\phi)} \ = a \ sec^{2} (\phi)$

$\sf \implies dL = a \ sec^{2}(\phi) d\phi$

$\sf \implies B = \displaystyle \dfrac{\mu_{o} i}{4 \pi} \int \dfrac{a \ sec^{2}\phi (d \phi) \ cos(\phi) \ cos^{2} \phi }{a^{2} }$

$\sf \implies B = \displaystyle \dfrac{\mu_{o} i}{4 \pi a} \int_{-\phi_{1}}^{\phi_{2}}} sec^{2} \phi \ d \phi \ cos \phi \ cos^{2} \phi$

$\sf \implies B = \displaystyle \dfrac{\mu_{o} i}{4 \pi a} \int_{-\phi_{1}}^{\phi_{2}}} cos(\phi) d\phi$

$\sf \implies B = \displaystyle \dfrac{\mu_{o} i}{4 \pi a} \bigg[sin(\phi) \bigg]_{\phi_{1}}}^{\phi_{2}}}$

$\sf \implies B = \displaystyle \dfrac{\mu_{o} i}{4 \pi a} \bigg[ sin(\phi_{2}}) \ - sin(-\phi_{1}) \bigg]$

$\sf \implies B = \displaystyle \dfrac{\mu_{o} i}{4 \pi a} \bigg[ sin(\phi_{2}}) \ + sin(\phi_{1}) \bigg]$