Question

A uniform magnetic field passes through a horizontal circular wire loop at an angle 16.2 ° from the normal to the plane of the loop. The magnitude of the magnetic field is 2.95 T , and the radius of the wire loop is 0.240 m . Find the magnetic flux Φ through the loop.

Answer 1

Answer:

Magnetic flux, $\phi=0.512\ Wb$

Explanation:

Given that,

The magnitude of the magnetic field is 2.95 T

Radius of the circular wire, r = 0.240 m

A uniform magnetic field passes through a horizontal circular wire loop at an angle 16.2 ° from the normal to the plane of the loop. The magnetic flux is given by the formula as :

$\phi=BA\ \cos\theta$

$\phi=2.95\times \pi \times (0.24)^2\ \cos(16.2)$

$\phi=0.512\ Wb$

So, the magnetic flux Φ through the loop is $\phi=0.512\ Wb$. Hence, this is the required solution.

Answer 2

Answer:

$\phi=0.512\;\;Weber$

Explanation:

Given,

Magnetic field $B=2.95\;\;T$

Radius $r=0.24\;\;m$

Angle $\theta=16.2^o$

Magnetic flux

$\phi=BAcos\theta\\\phi=2.95\times\pi (0.24)^2\times cos16.2^o\\\phi=0.512\;\;Weber$