A square loop of area $25 {\mathrm{cm}}^2$ has a resistance of $10 \Omega$. The loop is placed in uniform magnetic field of magnitude $40.0 \mathrm{T}$. 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 $1.0$ sec, will be

A massless square loop, of wire of resistance $10 \mathrm{\Omega }$, supporting a mass of $1 \mathrm{g}$, hangs vertically with one of its sides in a uniform magnetic field of ${10}^3 \mathrm{G}$, directed outwards in the shaded region. A dc voltage $V$ is applied to the loop. For what value of $V$, the magnetic force will exactly balance the weight of the supporting mass of $1 \mathrm{g}$?

(If sides of the loop $= 10 \mathrm{cm}, g=10 \mathrm{m} {\mathrm{s}}^{-2}$)

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As shown in the figure, a current of $2 \mathrm{A}$ flowing in an equilateral triangle of side $4\sqrt{3} \mathrm{cm}$ . The magnetic field at the centroid $O$ of the triangle is :

(Neglect the effect of earth’s magnetic field)

A current carrying rectangular loop $PQRS$ is made of uniform wire. The length $PR=QS=5 \mathrm{cm} \text{and} PQ=RS=100 \mathrm{cm}$. If ammeter current reading changes from $I$ to $2I$, the ratio of magnetic forces per unit length on the wire $PQ$ due to wire $RS$ in the two cases respectively $\left(f_{PQ}^I:f_{PQ}^{2I}\right)$ is:

A rod with circular cross-section area $2 {\mathrm{cm}}^2$ and length $40 \mathrm{cm}$ is wound uniformly with $400$ turns of an insulated wire. If a current of $0.4 \mathrm{A}$ flows in the wire windings, the total magnetic flux produced inside windings is $4\mathrm{\pi }\times {10}^{-6} \mathrm{Wb}$. The relative permeability of the rod is

(Given : Permeability of vacuum ${\mathrm{\mu }}_0=4\mathrm{\pi }\times {10}^{-7} \mathrm{N} {\mathrm{A}}^{-2}$)