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A total charge of $5 \times 10^{- 4} C$ passes through a 100 turn coil of resistance 200 $Ω$ , cross-sectional area 40cm² when it is rotated in a uniform field from a position where the plane of the coil is perpendicular to the field to a position where the coil’s plane is parallel to the field. Calculate B in millitesla.

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answer is 250.

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Detailed Solution

$I = \frac{d q}{d t} = \frac{ξ}{R} w h e r e ξ = - N \frac{d ϕ_{0}}{d t}$

$\Rightarrow \int_{0}^{Q} d q = \frac{N}{R} \int_{ϕ_{1}}^{ϕ_{2}} d ϕ_{0}$

$\Rightarrow | Q | = \frac{N}{R} (ϕ_{2} - ϕ_{1})$

$\Rightarrow Q = \frac{N}{R} [B A \cos (0^{0}) - B A \cos (\frac{π}{2})] = \frac{B A N}{R}$

$\Rightarrow B = \frac{R Q}{N A} = \frac{(200) (5 \times 10^{- 4})}{(100) (40 \times 10^{- 4})} = 0.25 T = 250 m T$

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