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Given:L=8 \mathrm{~cm}[ side of square coil ] \Rightarrow\left[L=\frac{8}{100} \mathrm{~m}\right]N=50 [number of turns]t=0.25 [time upto which coil is pulled]R=15 \Omega[ resiskance of coil ]I=12 \mathrm{~mA}[ current in coil ] \Rightarrow I=12 \times 10^{-3} \mathrm{~A}since, Area (A) of square coil =L^{2}\begin{aligned}\text { (initially) } & =\left(8 \times 10^{-2}\right)^{2} \mathrm{~m}^{2} \\& =64 \times 10^{-4} \mathrm{~m}^{2}\end{aligned}Since, magnetic field is perpendicular to the plane of coil , st, magneti field is parallel to the carce vecter which is normal to the plame of coil. Hence, B \| A\left(\theta=0^{8}\right) angle between \vec{B} and \vec{A}\begin{aligned}Q_{B}=(\text { magnetic fluc) } & \Rightarrow \vec{B} \cdot \vec{A} \\& =B A \cos 0^{\circ} \\\phi_{B} & =(B A)\end{aligned}Using the formula of induced cument\begin{aligned}I_{\text {ind }} & =-\frac{1}{R} \frac{d}{d t} \frac{\varnothing_{B}}{d t} \\I_{\text {ind }} & =-\frac{N}{R} \frac{d(B A)}{d t}\end{aligned}since, we pull coil out of uniform magnetic field, so B is constant and area get decrease or vanish from \left(64 \times 10^{-4} \mathrm{~m}^{2}\right) to zero in 0.2 sec as collis pulled quikly out of constant magnetic fieed.\begin{array}{r}\therefore \text { Iind }=-\frac{N}{R} B \frac{d A}{d t} \\\quad \text { Iind } R d t=N B d A\end{array}integrate both side with proper limits\begin{aligned}-\int_{0}^{0.2} \text { Iind } R d t & =\int_{64 \times 10^{-4}}^{0} N B d A \\+\int_{0}^{0.2} I_{\text {ind }} R d t & =t \int_{0}^{64 \times 10^{-4}} N B d A \\\text { Iind } R[t]_{0}^{0.2} & =N B[A]_{0}^{64 \times 10^{-4}} \mathrm{~N} \\\text { Iind } X R[0.2] & =N B\left[64 \times 10^{-4}\right] \\B & =\frac{0.2 \times 12 \times 15 \times 10^{-3}}{50 \times 64 \times 10^{-4}} \mathrm{~T} \\B & =0.11 T]\end{aligned}S0, The magnetiz field is 0.11 T.option (B) is correctoption (B) is correct ...