Question

A square loop of wire of side 5.0 cm is in a uniform magnetic field of 0.56 T. What is the magnetic flux in the loop when magnetic field vector is perpendicular to the face of the loop?

Answer 1

0.0014 weber

Step-by-step explanation:

We have given side of the square loop = 5 cm =0.05 meter

Area of the square loop A = 0.05×0.05=0.0025
`m^2`

It is given that magnetic field vector is perpendicular to the the face of the loop so the area vector and the magnetic field vector will be parallel to each other so angle between magnetic field and area will be zero

According to Lenz law magnetic flux
`\Phi =B.A=BAcos\Theta` here
`\Theta` is the angle between the magnetic field vector and the area vector

We have also given B = 0.56 T

So
`\Phi =0.56* 0.0025* cos0=0.0014\ weber`