Question

A coil of area 5${cm}^2$ with 20 turns is kept under themagnetic field of ${10}^3$ Gauss. Normal to the plane ofcoil makes an angle ${30}^{\circ }$ with the magnetic field. Theflux through the coil is

• A. $6.67\times {10}^{-4}Wb$
• B. $3.2\times {10}^{-5}Wb$
• C. $5.9\times {10}^{-4}Wb$
• D. $8.65\times {10}^{-4}Wb$

Answer 1

The correct option is D $8.65\times {10}^{-4}Wb$

$\text{Given}:$A=$5c{m}^2$, N=20, B=${10}^3$gauss, $\theta ={30}^o$

$\text{Solution}:$

$B={10}^3$ gauss

$={10}^3\times {10}^{4}T=0.1T$

$A=5c{m}^2=5\times {10}^{-4}{m}^2$

$\theta ={30}^{\circ }$

$\therefore$ Flux through the coil

$\varphi =NBA\cos \theta$

$=20\times 0.1\times 5\times {10}^{-4}\times \cos {30}^{\circ }$

$=10\times {10}^{-4}\times \frac{\sqrt{3}}{2}=5\sqrt{3}\times {10}^{-4}=8.65\times {10}^{-4}Wb$

$\text{Hence D is the correct option}$