Respuesta :
Answer:
Magnetic flux, [tex]\phi=0.512\ Wb[/tex]
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 :
[tex]\phi=BA\ \cos\theta[/tex]
[tex]\phi=2.95\times \pi \times (0.24)^2\ \cos(16.2)[/tex]
[tex]\phi=0.512\ Wb[/tex]
So, the magnetic flux Φ through the loop is [tex]\phi=0.512\ Wb[/tex]. Hence, this is the required solution.
Answer:
[tex]\phi=0.512\;\;Weber[/tex]
Explanation:
Given,
Magnetic field [tex]B=2.95\;\;T[/tex]
Radius [tex]r=0.24\;\;m[/tex]
Angle [tex]\theta=16.2^o[/tex]
Magnetic flux
[tex]\phi=BAcos\theta\\\phi=2.95\times\pi (0.24)^2\times cos16.2^o\\\phi=0.512\;\;Weber[/tex]
