A positive ion is moving with a speed of 5 10^5 m s ⁻¹ parallel to an infinitely long straight wire carrying a current of 20 A . The perpendicular distance between the ion and the wire is 5 cm . In order to keep the ion moving undeflected in its original path, a uniform electric field must be applied in the region. What is the required magnitude of this electric field?

Step-by-step solution

For the ion to move undeflected, the net force on it must be zero. This means the electric force must exactly balance the magnetic force in magnitude and be opposite in direction. The condition for undeflected motion is: qE = qvB E = vB The magnetic field B produced by a long straight wire at a distance r is: B = ₀ I 2 r Given: I = 20 A r = 5 cm = 0.05 m v = 5 10^5 m s ⁻¹ Calculating B : B = 4 10⁻⁷ 20 2 0.05 = 2 10⁻⁷ 20 0.05 = 8 10⁻⁵ T Now, calculate the required electric field E : E = vB = (5 10^5) (8 10⁻⁵) = 40 V