A circular disc of radius 0.2 meter is placed in a uniform magnetic field of induction (1/p)(Wb/m2) in such a way that its axis makes an angle of 60° with vecB. The magnetic flux (in Wb) linked with the disc is:

Question

A circular disc of radius 0.2 meter is placed in a uniform magnetic field of induction (1/p)(Wb/m2) in such a way that its axis makes an angle of 60° with vecB. The magnetic flux (in Wb) linked with the disc is: (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Here, B=(1/p)(Wb/m2) θ =60° Area normal to the plane of the disc = pr2 cos 60° = (pr2/2) Flux = B × normal area = (0.2 ×0.2/2) = 0.02Wb

Q. A circular disc of radius $0.2$ meter is placed in a uniform magnetic field of induction $\frac{1}{p} (Wb / m^{2})$ in such a way that its axis makes an angle of $6 0^{\circ}$ with $B$ . The magnetic flux (in Wb) linked with the disc is:

Here, $B = \frac{1}{p} (Wb / m^{2}) θ = 6 0^{\circ}$
Area normal to the plane of the disc
$= p r^{2}$ cos $6 0^{\circ} = \frac{p r ^{2}}{2}$
Flux $= B \times$ normal area
$= \frac{0.2 \times 0.2}{2} = 0.02 Wb$

Q. A circular disc of radius 0.2 meter is placed in a uniform magnetic field of induction p1​(Wb/m2) in such a way that its axis makes an angle of 60∘ with B. The magnetic flux (in Wb) linked with the disc is:

Here, B=p1​(Wb/m2)θ=60∘ Area normal to the plane of the disc =pr2 cos 60∘=2pr2​ Flux =B× normal area =20.2×0.2​=0.02Wb

Q. A circular disc of radius $0.2$ meter is placed in a uniform magnetic field of induction $\frac{1}{p} (Wb / m^{2})$ in such a way that its axis makes an angle of $6 0^{\circ}$ with $B$ . The magnetic flux (in Wb) linked with the disc is:

Here, $B = \frac{1}{p} (Wb / m^{2}) θ = 6 0^{\circ}$
Area normal to the plane of the disc
$= p r^{2}$ cos $6 0^{\circ} = \frac{p r ^{2}}{2}$
Flux $= B \times$ normal area
$= \frac{0.2 \times 0.2}{2} = 0.02 Wb$