Question

A circular loop of wire has an area of 0.27 m^2 . It is tilted by 45° with respect to a uniform 0.38T magnetic field. What is the magnetic flux through the loop?

Answer 1

The magnetic flux is 0.0725 Wb

Solution:

Area of the wirer,
`A_(w) = 0.27 m^(2)`

Magnetic field associated with the wire, B = 0.38 T

Angle with the magnetic field,
`\theta = 45^(\circ)`

Now, we know that the magnetic flux is given by:

`\phi_(w) = \vec{B}.\vec{A_(w)}`

`\phi_(w) = BA_(w)cos\theta`

Now, using the above eqn:

`\phi_(w) = 0.38* 0.27cos45^(\circ)`

`\phi_(w) = 0.38* 0.27cos45^(\circ) = 0.0725 Wb`