A square loop of area $$25 \mathrm{~cm}^2$$ has a resistance of $$10 \Omega$$. This loop is placed in a uniform magnetic field of magnitude $$40 \mathrm{~T}$$. The plane of loop is perpendicular to the magnetic field. The work done in pulling the loop out of the magnetic field slowly and uniformly in one second, will be

Two conducting circular loops of radii '$$R_1$$' and '$$R_2$$' are placed in the same plane with their centres coinciding. If $$R_1>R_2$$, the mutual inductance $$M$$ between them will be directly proportional to

If current '$$I$$' is flowing in the closed circuit with collective resistance '$$R$$', the rate of production of heat energy in the loop as we pull it along with a constant speed '$$\mathrm{V}$$' is ( $$\mathrm{L}=$$ length of conductor, $$\mathrm{B}=$$ magnetic field)

Two coils $$\mathrm{A}$$ and $$\mathrm{B}$$ have mutual inductance 0.008 $$\mathrm{H}$$. The current changes in the coil A, according to the equation $$\mathrm{I}=\mathrm{I}_{\mathrm{m}} \sin \omega \mathrm{t}$$, where $$\mathrm{I}_{\mathrm{m}}=5 \mathrm{~A}$$ and $$\omega=200 \pi ~\mathrm{rad} ~\mathrm{s}^{-1}$$. The maximum value of the e.m.f. induced in the coil $$B$$ in volt is