Answer:
The resulting magnetic force on the wire is -1.2kN
Explanation:
The magnetic force on a current carrying wire of length 'L' with current 'I' in a magnetic field B is
F = I (L*B)
Finding (L * B) , where L = (2, 0, 0)m , B = (30, -40, 0)
L x B = = (0, 0, -80)
we can now solve
F = I (L x B) = I (-80)
F = -1200 kmN
F = -1200 kN * 10⁻³
F = -1.2kN
Answer:
Approximately (given that the magnitude of this charge is
.)
Explanation:
If a charge of magnitude is placed in an electric field of magnitude
, the magnitude of the electrostatic force on that charge would be
.
The magnitude of this charge is . Apply the unit conversion
:
.
An electric field of magnitude would exert on this charge a force with a magnitude of:
.
Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.
Answer:
Explanation:
For this exercise we must use the principle of conservation of energy
starting point. The proton very far from the nucleus
Em₀ = K = ½ m v²
final point. The point where the proton is stopped (v = 0)
Em_f = U = q V
where the potential is
V = k Ze / r²
Let us consider that all the charge of the nucleus is in the center, therefore r is the distance from this point to the proton that is approaching
Energy is conserved
Em₀ = Em_f
½ m v² = e ()
with this expression we can find the closest approach distance (r)