SATHEE: Electromagnetic Induction

Electromagnetic Induction - Result Question 13

13. A circular disc of radius 0.2 meter is placed in a uniform magnetic field of induction $\frac{1}{p} (W b / m^{2})$ in such a way that its axis makes an angle of $60^{\circ}$ with $\vec{B}$ . The magnetic flux linked with the disc is:

[2008]

(a) $0.02 W b$

(b) $0.06 W b$

(d) $0.01 W b$

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Answer:

Correct Answer: 13. (a)

Solution:

(a) Here, $B = \frac{1}{p} (W b / m^{2})$

$θ = 60^{\circ}$

Area normal to the plane of the disc

$= p r^{2} \cos 60^{\circ} = \frac{p r^{2}}{2}$

Flux $= B \times$ normal area

$= \frac{{0.2}^{'} 0.2}{2} = 0.02 W b$

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Electromagnetic Induction - Result Question 13

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

Answer:

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13. A circular disc of radius 0.2 meter is placed in a uniform magnetic field of induction $\frac{1}{p} (W b / m^{2})$ in such a way that its axis makes an angle of $60^{\circ}$ with $\vec{B}$ . The magnetic flux linked with the disc is: