The correct Biot-Savart law in vector form is

An electron is moving in a circle of radius $r$ in a uniform magnetic field $B$. Suddenly, the field is reduced to $\frac{B}{2}$. The radius of the circular path now becomes

A charge $q$ is accelerated through a potential difference $V$. It is then passed normally through a uniform magnetic field, where it moves in a circle of radius $r$. The potential difference required to move it in a circle of radius $2 r$ is

A cyclotron's oscillator frequency is 10 MHz and the operating magnetic field is 0.66 T . If the radius of its dees is 60 cm , then the kinetic energy of the proton beam produced by the accelerator is