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A conducting circular loop having a radius of 1.0 cm, is placed perpendicular to a magnetic field of 0.50 T .It is removed from the field in 0.50 s. The average emf produced in the loop during this time is $x π \times 10^{- 4} V$ then find out value of x

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$0.5 π \times 10^{- 3} V$

$π \times 10^{- 3} V$

$2 π \times 10^{- 4} V$

$π \times 10^{- 4} V$

answer is A.

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Detailed Solution

$ε = - \frac{Δϕ}{Δt} = - \frac{(ϕ_{2} - ϕ_{1})}{Δt} = \frac{ϕ_{1} - ϕ_{2}}{Δt} = \frac{BA - 0}{Δt} = \frac{0 .5 \times π {(1 \times 10^{- 2})}^{2}}{0 .5} = π \times 10^{- 4} V$

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