Answer :
Answer:
Magnetic flux, [tex]\phi=2.22\times 10^{-3}\ Wb[/tex]
Explanation:
It is given that,
Magnitude of the magnetic field, B = 0.078 T
Radius of circular loop, r = 0.1 m
The field is oriented at an angle of θ = 25° with respect to the normal to the surface. The magnetic flux through the surface is given by :
[tex]\phi=BA\ cos\theta[/tex]
[tex]\phi=0.078\times \pi \times (0.1)^2\ cos(25)[/tex]
[tex]\phi=0.00222\ Wb[/tex]
or
[tex]\phi=2.22\times 10^{-3}\ Wb[/tex]
So, the magnitude of magnetic flux through the surface is [tex]2.22\times 10^{-3}\ Wb[/tex]. Hence, this is the required solution.
Answer:
0.00221Tm^2
Explanation:
B( magnitude of magnetic field )=0.078T
r=0.1m
theta=25°
magnetic flux=?
Area=πr^2
=3.14×0.1^2
=0.0314m^2
Magnetic flux= BACostheta
=0.078×0.0314×Cos25°
=0.00221Tm^2 /0.00221wb