Question

An ink drop with charge q = 3 ✕ 10-9 c is moving in a region containing both an electric field and a magnetic field. the strength of the electric field is 9 105 n/c and the strength of the magnetic field is 1.3 t. at what speed must the particle be moving perpendicular to the magnetic field so that the magnitudes of the electric and magnetic forces are equal?

Answer 1

Since charge is moving inside both type of field

there will exist both forces on it

now

$F = q(v x B) + qE$

If net force on moving charge is zero

$0 = qvBsin90 + qE$

$qvB = qE$

hence we will have speed

$v = \frac{E}{B}$

$v = \frac{9 * 10^5}{1.3}$

$v = 6.92 * 10^5 m/s$