Two long parallel wires carrying currents $$8 \mathrm{~A}$$ and $$15 \mathrm{~A}$$ in opposite directions are placed at a distance of $$7 \mathrm{~cm}$$ from each other. A point '$$\mathrm{P}$$' is at equidistant from both the wires such that the lines joining the point to the wires are perpendicular to each other. The magnitude of magnetic field at point '$$\mathrm{P}$$' is $$(\sqrt{2}=1.4) ( \mu_0=4 \pi \times 10^{-7}$$ SI units)

Electron of mass '$$\mathrm{m}$$' and charge '$$\mathrm{q}$$' is travelling with speed '$$v$$' along a circular path of radius '$$R$$', at right angles to a uniform magnetic field of intensity '$$B$$'. If the speed of the electron is halved and the magnetic field is doubled, the resulting path would have radius

10 A current is flowing in two straight parallel wires in the same direction. Force of attraction between them is $$1 \times 10^{-3} \mathrm{~N}$$. If the current is doubled in both the wires the force will be

The magnetic field at a point $$\mathrm{P}$$ situated at perpendicular distance '$$R$$' from a long straight wire carrying a current of $$12 \mathrm{~A}$$ is $$3 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^2$$. The value of '$$\mathrm{R}$$' in $$\mathrm{mm}$$ is $$\left[\mu_0=4 \pi \times 10^{-7} \mathrm{~Wb} / \mathrm{Am}\right]$$