Answer:
the magnitude of the magnetic force on the wire segment is 6.03 x 10⁻⁶ N
Explanation:
Given;
length of the conductor, L = 1 cm = 0.01 m
current carried by the solenoid, I₁ = 15 A
radius of the solenoid, r = 4 cm
number of turns per length of the solenoid, n = 800 turns/m
current carried by the solenoid, I₂ = 40 mA = 0.04 A
The magnetic field of the solenoid is calculated as;
B = μnI₂
where;
μ is the permeability of free space = 4π x 10⁻⁷ Tm/A
B = ( 4π x 10⁻⁷) x (800) x (0.04)
B = 4.022 x 10⁻⁵ T
The magnitude of the magnetic force on the wire segment is calculated as;
F = BI₁L sinθ
where
θ is the angle made by the wire segment against the solenoid = 90⁰
F = (4.022 x 10⁻⁵) x (15) x (0.01) x sin(90)
F = 6.03 x 10⁻⁶ N
Therefore, the magnitude of the magnetic force on the wire segment is 6.03 x 10⁻⁶ N
