$A$ light charged particle is revolving in a circle of radius $r$ due to the electrostatic attraction of a static heavy particle with an opposite charge. How does the magnetic field $B$ at the centre of the circle, produced by the moving charge, depend on $r$?

An electron (mass $9 \times 10^{-31} \ kg$ and charge $1.6 \times 10^{-19} \ C$) moving with speed $v = c/100$ $(c = 3 \times 10^8 \ ms^{-1})$ is injected into a magnetic field $\vec{B}$ of magnitude $9 \times 10^{-4} \ T$ perpendicular to its direction of motion. We wish to apply a uniform electric field $\vec{E}$ together with the magnetic field so that the electron does not deflect from its path. Then:

One Tesla is equal to

As the temperature of the hot junction increases, the thermo $e.m.f.$

The magnetic field intensity at the point $O$ of a loop with current $i$,whose shape is illustrated below is

Two long current-carrying thin wires,both with current $I$,are held by insulating threads of length $L$ and are in equilibrium as shown in the figure,with threads making an angle $\theta$ with the vertical. If the wires have mass $\lambda$ per unit length,then the value of current $I$ is ($g =$ gravitational acceleration).