Respuesta :
Answer:
(a). The strength of the magnetic field is 0.1933 T.
(b). The magnetic flux through the loop is zero.
Explanation:
Given that,
Radius = 11.9 cm
Magnetic flux [tex]\phi=8.60\times10^{-3}\ T m^2[/tex]
(a). We need to calculate the strength of the magnetic field
Using formula of magnetic flux
[tex]\phi=BA\cos\theta[/tex]
[tex]\phi=BA\cos0[/tex]
[tex]\phi=BA[/tex]
[tex]B=\dfrac{\phi}{A}[/tex]
[tex]B=\dfrac{\phi}{\pi r^2}[/tex]
Put the value into the formula
[tex]B=\dfrac{8.60\times10^{-3}}{\pi\times(11.9\times10^{-2})^2}[/tex]
[tex]B=0.1933\ T[/tex]
(b). If the magnetic field is directed parallel to the plane of the loop,
We need to calculate the magnetic flux through the loop
Using formula of flux
[tex]\phi=BA\cos\theta[/tex]
Here, [tex]\theta=90^{\circ}[/tex]
[tex]\phi=BA\cos90[/tex]
[tex]\phi=0[/tex]
Hence, (a). The strength of the magnetic field is 0.1933 T.
(b). The magnetic flux through the loop is zero.
Answer:
Explanation:
radius of loop, r = 11.9 cm = 0.119 m
magnetic flux, Ф = 8.6 x 10^-3 T m²
(a) As the field is perpendicular to the plane of the loop so teh angle between area vector and the magnetic field vector is zero.
Ф = B x A x Cosθ
where, B is the strength of magnetic field, A be the area of the loop and θ is the angle between the area vector and the magnetic field vector.
8.6 x 10^-3 = B x 3.14 x 0.119 x 0.119
B = 0.193 Tesla
(b) the magnetic field is parallel to the plane of loop, so θ = 90°, so the magnetic flux is
Ф = B x A x Cos 90
Ф = 0 T m²
