Question

A circular loop of radius 11.9 cm is placed in a uniform magnetic field. (a) If the field is directed perpendicular to the plane of the loop and the magnetic flux through the loop is 8.60 ✕ 10−3 T · m2, what is the strength of the magnetic field? T (b) If the magnetic field is directed parallel to the plane of the loop, what is the magnetic flux through the loop? T · m2

Answer 1

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

Answer 2

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 $\phi=8.60\times10^{-3}\ T m^2$

(a). We need to calculate the strength of the magnetic field

Using formula of magnetic flux

$\phi=BA\cos\theta$

$\phi=BA\cos0$

$\phi=BA$

$B=\dfrac{\phi}{A}$

$B=\dfrac{\phi}{\pi r^2}$

Put the value into the formula

$B=\dfrac{8.60\times10^{-3}}{\pi\times(11.9\times10^{-2})^2}$

$B=0.1933\ T$

(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

$\phi=BA\cos\theta$

Here, $\theta=90^{\circ}$

$\phi=BA\cos90$

$\phi=0$

Hence, (a). The strength of the magnetic field is 0.1933 T.

(b). The magnetic flux through the loop is zero.