Question

An electron (e = 1.6 x 10-19 C) is traveling at 4.00 x 107 m/s due North in a horizontal plane through a point where the earth’s magnetic field has a component to the North of 2.00 x 10-5 T and a downward (in to the earth) of 5.00 x 10-5 T. Calculate the magnetic force (magnitude and direction) on the electron and its acceleration. Me = 9.11 x 10-31 kg

Answer 1

Step-by-step explanation:

Force on the electron due to magnetic field 's north component will be zero because both velocity of electron and direction of magnetic field are same .

Force due to downward component of magnetic field

= B q v where B is magnetic field , q is charge moving and v is velocity of charge

F = 5 .00 x 10⁻⁵ x 1.6 x 10⁻¹⁹ x 4 x 10⁷

= 32 x 10⁻¹⁷ N

acceleration = F / m where m is mass of electron

= 32 x 10⁻¹⁷ / 9.11 x 10⁻³¹

= 3.5 x 10¹⁴ m/s²