A 1-cm long wire carrying 15 A is inside a solenoid 4 cm in radius with 800 turns/m carrying a current of 40 mA. The wire segment is oriented perpendicularly to the axis of the solenoid. What is the magnitude of the magnetic force on this wire segment in N?

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

reserve is the answer

you get 2.3=1.21*\frac{a}2 or 4.6=1.21a or a=3.802 (b)

explanation: