Question

PLS give the correct solution by checking the answer.​

Answer 1

$\Huge\underline{\underline{\bf \red{Question\:18}:}}$

$\Large\underline{\underline{\sf Given:}}$

• Current I is flowing in a given conductor

• Radius of curved part is r

• Lenght of straight wire is infinity.

$\Large\underline{\underline{\sf To\:Find:}}$

• Magnetic Field at the centre O

$\Large\underline{\underline{\sf Formula\:Used:}}$

Magnetic Field for circle is

${\boxed{\boxed{\sf \pink{B=\dfrac{\mu_oI}{2r}}}}}$

$\Large\underline{\underline{\sf Solution:}}$

Magnetic Field at the centre O

${\boxed{\sf B=B_{FED}+B_{DC}+B_{FG}}}$

Magnetic field for ¾ of circle is :

$\implies{\sf B_{FED}=\dfrac{\mu_oI}{r}×\dfrac{3}{4}{\hat{k}} }$

$\implies{\sf B_{DC}=0}$

$\implies{\sf B_{FG}=\dfrac{\mu_oI}{4πd}[sin\theta_1+sin\theta_2]}$

$\implies{\sf \dfrac{\mu_oI}{4πd}[sin0°+sin90°]}$

$\implies{\sf \dfrac{\mu_oI}{4πd}[0+1] }$

$\implies{\sf \dfrac{\mu_oI}{4πr}}$

Magnetic field at Centre O

$\large{\sf B=\left[\dfrac{3}{4}×\dfrac{\mu_oI}{2π}+\dfrac{\mu_oI}{4πr}\right]{\hat{k}} }$

$\implies{\sf B=\dfrac{\mu_oI}{4r}\left[\dfrac{3}{2}+\dfrac{1}{π}\right]{\hat{k}} }$

$\implies{\sf B=\dfrac{\mu_oI}{4πr}\left[\dfrac{3π}{2}+1\right]}$

$\Large\underline{\underline{\sf Answer:}}$

Magnetic field at Centre O is ${\sf B=\dfrac{\mu_oI}{4πr}\left[\dfrac{3π}{2}+1\right]}$

$\Huge\underline{\underline{\bf \red{Question\:19}:}}$

$\Large\underline{\underline{\sf To\:Find:}}$

Magnetic Induction in the given figure .

$\Large\underline{\underline{\sf Solution:}}$

Magnetic Induction due to semi-circular part

${\boxed{\boxed{\sf \pink{B=\dfrac{\mu_oI}{4r}}}}}$

•°• Magnetic Induction for Radius $\sf{R_1}$ and radius $\sf{R_2}$ will be

$\implies{\sf B=\dfrac{\mu_oI}{4}\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)}$

$\Large\underline{\underline{\sf Answer:}}$

⛬ Magnetic Induction of the given figure is ${\sf B=\dfrac{\mu_oI}{4}\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)}$