Question

An electron is accelerated through 2400 V from rest and then enters a region in which there is a uniform 1.70 T magnetic field. What are (a) the maximum and (b) the minimum magnitudes for the magnetic force this charge can experience

Answer 1

Answer:

Explanation:

Let v be the velocity acquired by electron in electric field

V q = 1/2 m v²

V is potential difference applied on charge q , m is mass of charge , v is velocity acquired

2400 x 1.6 x 10⁻¹⁹ = .5 x 9.1 x 10⁻³¹ x v²

v² = 844 x 10¹²

v = 29.05 x 10⁶ m /s

Maximum force will be exerted on moving electron when it moves perpendicular to magnetic field .

Maximum force = Bqv , where B is magnetic field , q is charge on electron and v is velocity of electron

= 1.7 x 1.6 x 10⁻¹⁹ x 29.05 x 10⁶

= 79.02 x 10⁻¹³ N .

Minimum force will be zero when electron moves along the direction of magnetic field .