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A circular disc of radius 0.2 m is placed in a uniform magnetic field of induction \mathrm {\mathrm{n}\left(\frac{1}{\pi}\right)Wbm^{-2}}  in such a way that its axis makes an angle of 60° with B. The magnetic flux linked with the disc is

Option: 1

0.02 Wb


Option: 2

0.06 Wb


Option: 3

0.08 Wb


Option: 4

0.01 Wb


Answers (1)

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\mathrm{\text { Here, } \mathrm{r}=0.2 \mathrm{~m}, \mathrm{~B}=\frac{1}{\pi} \mathrm{Wbm}^{-2}, \theta=60^{\circ}}

\mathrm{\therefore \text { Magnetic flux, } \phi=\mathrm{BA} \cos \theta=\mathrm{B}\left(\pi \mathrm{r}^2\right) \cos \theta=\frac{1}{\pi} \pi(0.2)^2 \cos 60^{\circ}=0.02 \mathrm{~Wb}}

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Divya Prakash Singh

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