Question

A coil of area 5 $c{m}^2$ with $20$ turns is kept under the magnetic field of ${10}^3$ Gauss. Normal to the plane of coil makes an angle ${30}^o$ with the magnetic field. The flux 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 $8.65\times {10}^{-4}Wb$

Given: N = 20
$B={10}^3$ gauss
$={10}^3\times {10}^{-4}T=0.1T$
$A=5c{m}^2=5\times {10}^{-4}{m}^2$
$\theta ={30}^o$
$\therefore$ Flux through the coil
$\varphi =NBAcos\theta$
$=20\times 0.1\times 5\times {10}^{-4}\times cos{30}^o$
$=10\times {10}^{-4}\times \frac{\sqrt{3}}{2}=5\sqrt{3}\times {10}^{-4}=8.65\times {10}^{-4}wb$