Question A square coil of wire with side 8.0 cm and 50 turns sits in a uniform magnetic field that is perpendicular to the plane of the coil. The coil is pulled quickly out of the magnetic field in 0.2 s. If the resistance of the coil is 15 ohm and a current of 12 mA is induced in the coil, calculate the value of the magnetic field. O 5.6T O 0.11 T O 7.5 x 10-T O 1.4 T 9.1 T

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 ...