A conducting circular loop made of a thin wire has area 3.5 × 10 ^...

It is given that, the magnetic field,

$B = 0.4\sin \left(50\pit\right)$

and the area of the given circular conducting loop,

$A = 3.5 \times {10}^{- 2}{m}^{2}$

Magnetic flux,

$ϕ = BA . . . ( 1 )$

Then, induced emf,

$ε = \left|\dfrac{dϕ}{dt}\right| . . . ( 2 )$

When resistance in the circuit is $R$, the differential charge accumulated, based on current $i$,

$dQ = idt = \dfrac{ε}{R}dt = \dfrac{1}{R}dϕ ⇒ ∆ Q = \dfrac{1}{R} ∫ d\phi = \dfrac{1}{R} ∆ ϕ$

Substituting given values and using equation $( 1 )$ and $( 2 )$,
$∆ Q = \dfrac{1}{10}\left[0.4\sin \left(50\pi \times 10 \times {10}^{- 3}\right) - 0\right] \times 3 . 5 \times {10}^{- 2} ⇒ ∆ Q = \dfrac{0.4 \times 3 . 5 \times {10}^{- 2}}{10} = 1 . 4 \times {10}^{- 3}C$

$⇒ ∆ Q = 1 . 4mC$