The magnetic field inside a current carrying toroidal solenoid is $$0.2 \mathrm{~mT}$$. What is the magnetic field inside the toroid if the current through it is tripled and radius is made $$\frac{1}{3}^{\text {rd}}$$ ?

A, B and C are three parallel conductors of equal lengths carrying currents $$\mathrm{I}, \mathrm{I}$$ and $$2 \mathrm{I}$$ respectively. Distance between A and B is '$$x$$' and that between B and C is also '$$x$$'. $$F_1$$ is the force exerted by conductor $$\mathrm{B}$$ on $$\mathrm{A}$$. $$\mathrm{F}_2$$ is the force exerted by conductor $$\mathrm{C}$$ on $$\mathrm{A}$$. Current $$\mathrm{I}$$ in $$\mathrm{A}$$ and $$\mathrm{I}$$ in $$\mathrm{B}$$ are in same direction and current $$2 \mathrm{I}$$ in $$\mathrm{C}$$ is in opposite direction. Then

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.]