Question

A square coil of $0.01m^2$ area is placed perpendicular to the uniform magnetic field of ${10}^{3}weber/mete{r}^2$. The magnetic flux linked with the coil is :-

• A. $10$ weber
• B. ${10}^{-5}$ weber
• C. Zero
• D. $100$ weber

Answer 1

Answer: (A)

$10$ weber

Given: Area of coil,$A=0.01m^2={10}^{-2}{m}^2$

magnetic field, $B={10}^{3}Wb/m^2$

To Find: magnetic flux, $\varphi =?$

Solution: As we know that,

magnetic flux($\varphi$) linked with a coil is equal to the multiplication of magnetic field($B$) and area of the coil($A$).

i.e. $\varphi =B\ast A$

$\varphi ={10}^3\ast {10}^{-2}$

$\varphi =10weber$

hence,

The correct opt: A