What should be the minimum magnitude and direction of the magnetic field that must be produced at the equator of earth so that a proton may go round the earth with a speed of $1.0 \mathrm{x} {10}^7 {\mathrm{ms}}^{-1}$? Earth’s radius is$6.4 \mathrm{x} {10}^6 \mathrm{m}.$

A current-carrying conductor has$8.0 \times {10}^{22}$ free electrons per metre length and their mean drift velocity is$8.0 \times {10}^{-5} {\mathrm{ms}}^{-1}$ If a magnetic field of$0.10 \mathrm{T}$ be applied perpendicular to the conductor, then calculate

(i) current in the conductor, 

A current-carrying conductor has $8.0 \times {10}^{22}$ tree electrons per metre length and their mean drift velocity is $8.0 \times {10}^{-5}{\mathrm{ms}}^{-1}$ If a magnetic field of $0.10 \mathrm{T}$ be applied perpendicular to the conductor, then calculate 

(ii) force on an electron and

A current-carrying conductor has $8.0 \times {10}^{22}$ free electrons per metre length and their mean drift velocity is $8.0 \mathrm{x} {10}^{-5} {\mathrm{ms}}^{-1}$. If a magnetic field of $0.10 \mathrm{T}$ be applied perpendicular to the conductor, then calculate the force imposed per metre length of the conductor.