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

The magnetic field due to a current in a straight wire segment of length L at a point on its perpendicular bisector at a distance r(r >> L)

The magnets of two suspended coil galvanometers are of the same strength so that they produce identical uniform magnetic fields in the region of the coils. The coil of the first one is in the shape of a square of side a and that of the second one is circular of radius $${a \over {\sqrt \pi }}$$. When the same current is passed through the coils, the ratio of the torque experienced by the first coil to that experienced by the second one is

A proton is moving with a uniform velocity of $${{{10}^6}}$$ ms^$$-$$1 along the Y-axis, under the joint action of a magnetic field along Z-axis and an electric field of magnitude 2 $$ \times $$ 10⁴ Vm^$$-$$1 along the negative X-axis. If the electric field is switched off, the proton starts moving in a circle. The radius of the circle is nearly (given : $${e \over m}$$ ratio for proton $$ \approx $$ 10⁸ Ckg^$$-$$1)