An electron with kinetic energy 5 eV enters a region of uniform magnetic field of 3 μ T perpendicular to its direction. An electric field E is applied perpendicular to the direction of velocity and magnetic field. The value of E, so that electron moves along the same path, is __________ NC⁻¹.
(Given, mass of electron =9 × 10⁻³¹ kg, electric charge =1.6 × 10⁻¹⁹ C)

Solution

Physics sir · 4 steps

For a straight path, electric push must cancel magnetic push, so all you need is the speed: E = vB.

1. Balance the forces. The electron keeps its straight path only if the net force on it is zero. The electric force eE must cancel the magnetic force evB, so eE = evB and the charge cancels.

   E = vB, v = √(2K/m)

2. Find the speed. K = 5 × 1.6 × 10⁻¹⁹ = 8 × 10⁻¹⁹ J. Then v = √((2 × 8 × 10⁻¹⁹)/(9 × 10⁻³¹)) = √(16/9 × 10¹²) = 4/3 × 10⁶ m s⁻¹.
3. Put in numbers. E = 4/3 × 10⁶ × 3 × 10⁻⁶ N C⁻¹, where B = 3 μT = 3 × 10⁻⁶ T.
4. So. E = 4 N C⁻¹, so the answer is 4.

The common slip: 2.83. Dropping the 2 in K = 1/2mv² gives v ≈ 9.43 × 10⁵ m s⁻¹ and E ≈ 2.83, not 4.

The idea to keep. Crossed E and B act as a velocity selector: a charge goes straight only when v = E/B. Get v from the kinetic energy, and convert eV to joules first.

Answer from the NTA final answer key. Solution by PrepWiser.