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
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
because velocity is a vector quantity which has both magnitude as well as a specific direction and here the meteor's direction is specified in the statement hence we conclude that this statement describes meteor's velocity as well as speed too.
