Magnetic moment of revolving electron of charge (e) and mass (m) in terms of angular momentum (L) of electron is :

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