Question

You wish to hit a target from several meters away with a charged coin having a mass of 4.7 g and a charge of +2400 μC . The coin is given an initial velocity of 14.0 m/s , and a downward, uniform electric field with field strength 28.0 N/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:

$B = 3.37 T$

Direction = perpendicular to velocity and into the plane of the motion

Explanation:

As the coin is projected directly towards the target so due to electric field present in the region the coin will have tendency to deflect in the direction of electric field

Now in order to hit the target we need to counter balance the electric field force and gravitational force with the magnetic field force

So here we can say

$qE + mg = qvB$

here we know that

$v = 14 m/s$

$Q = 2400 \mu C$

E = 28 N/C

m = 4.7 g

now from above equation

$(2400 \times 10^{-6})(28) + (4.7 \times 10^{-3})(9.81) = (2400 \times 10^{-6})(14) B$

$0.0672 + 0.0461 = 0.0336 B$

$B = 3.37 T$

Direction = perpendicular to velocity and into the plane of the motion