A proton is moving in the x - y plane at an angle of 45^ to the positive x -axis with a kinetic energy of 10 keV . A uniform magnetic field B = 50 k mT exists in the region. The electric field vector E required to keep the proton moving undeflected in a straight line is: (Given: mass of proton = 1.6 10⁻²⁷ kg , charge of proton = 1.6 10⁻¹⁹ C )

Step-by-step solution

First, find the speed of the proton from its kinetic energy: K = 10 keV = 10 10³ 1.6 10⁻¹⁹ J = 1.6 10⁻¹⁵ J Using K = 1 2 mv² : v = 2K m = 2 1.6 10⁻¹⁵ 1.6 10⁻²⁷ = 2 10¹² = 2 10⁶ m s ⁻¹ The proton is moving in the x - y plane at 45^ to the positive x -axis. Its velocity vector is: v = v (45^ ) i + v (45^ ) j = ( 2 10⁶) ( 1 2 i + 1 2 j ) = 10⁶( i + j ) m s ⁻¹ For the proton to move undeflected, the net Lorentz force must be zero: F _ net = q( E + v B ) = 0 E = -( v B ) Given B = 50 10⁻³ k T : v B = [10⁶( i + j )] [50