Question

You wish to hit a target from several meters away with a charged coin
having a mass of 4.25 g and a charge of +2500 µC. The coin is given
an initial velocity of 12.8m/s, and a downward, uniform electric field
with field strength 27.5N/C exists throughout the region. If you aim
directly at the target and fire the coin horizontally, what magnitude
and direction of uniform magnetic field are needed in the region for the
coin to hit the target?

Answer 1

Answer:

B = 3.45 T

Explanation:

Given data:

mass of coin = 4.25 g

charge $+2500 \mu C$

initial velocity 12.8 m/s

Electric field strength 27.5 N.C

The force due to magnetic field must cancel force due to electric field  and gravitational force

qvB = qE + mg $B = \frac{1}{v} [ E +  \frac{mg}{q}]$

$=\frac{1}{12.8}[27.5+ \frac{4.25\times 10^{-3} 9.8}{2500\times 10^{-6}}]$

B = 3.45 T