The magnetic flux near the axis and inside the air core solenoid of length $$60 \mathrm{~cm}$$ carrying current '$$\mathrm{I}$$' is $$1.57 \times 10^{-6} \mathrm{~Wb}$$. Its magnetic moment will be $$\left[\mu_0=4 \pi \times 10^{-7}\right.$$, SI unit and crosssectional area is very small as compared to length of solenoid.]

A charge moves with velocity '$$V$$' through electric field $$(E)$$ as well as magnetic field (B). then the force acting on it is

A long solenoid carrying a current produces a magnetic field B along its axis. If the number of turns per $$\mathrm{cm}$$ is doubled and the current is made $$\left(\frac{1}{3}\right)^{\text {rd }}$$ then the new value of the magnetic field will be

A metal conductor of length $$1 \mathrm{~m}$$ rotates vertically about one of its ends at an angular velocity of $$5 \mathrm{~rad} / \mathrm{s}$$. If horizontal component of earth's magnetic field is $$0.2 \times 10^{-4} \mathrm{~T}$$, then the e.m.f. developed between the two ends of the conductor is