Question

A square loop of side 4m is placed in magnetic field of 10T, surface of loop is making an angle with the field 60°. What will be the flux?​

Answer 1

Given:

A square loop of side 4m is placed in magnetic field of 10T.

To find:

Magnetic flux ?

Calculation:

• Since the surface makes 60° with field intensity vector, the area vector will make (90°-60°) = 30° with the field intensity vector.

• Let flux be $\phi$

$\phi = B \times A \times \cos( \theta)$

$\implies \phi = 10 \times (4 \times 4) \times \cos( {30}^{ \circ} )$

$\implies \phi = 160\times \cos( {30}^{ \circ} )$

$\implies \phi = 160\times \dfrac{ \sqrt{3} }{2}$

$\implies \phi = 80 \sqrt{3} \: W$

So, magnetic flux is 80√3 Weber.