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

`\phi=0.512\;\;Weber`

Step-by-step 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*\pi (0.24)^2* cos16.2^o\\\phi=0.512\;\;Weber`

Answer 2

Magnetic flux,
`\phi=0.512\ Wb`

Step-by-step 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* \pi * (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.