Asolenoid having n turns and carrying a current of 2.000 a has a length of 34 00 cm. if the magnitude of the magnetic field generated at the center of the solenoid is 9.000 mt what is the value of n? (μo = 4π x10^-7 t. m/a) a) 2318 b) 1218 c) 860.0 d) 3183 e) 1591

B) 1218

Explanation:

N = Total number of turns in the solenoid

L = length of the solenoid = 34.00 cm = 0.34 m

B = magnetic field at the center of the solenoid = 9 mT = 9 x 10⁻³ T

i = current carried by the solenoid = 2.000 A

Magnetic field at the center of the solenoid is given as

N = 1218

The value of N is about B) 1218

Let's recall magnetic field strength from current carrying wire and from center of the solenoid as follows:

B = magnetic field strength from current carrying wire (T)

μo = permeability of free space = 4π × 10⁻⁷ (Tm/A)

I = current (A)

d = distance (m)

B = magnetic field strength at the center of the solenoid (T)

μo = permeability of free space = 4π × 10⁻⁷ (Tm/A)

I = current (A)

N = number of turns

L = length of solenoid (m)

Let's tackle the problem now !

Given:

Current = I = 2000 A

Length = L = 34.00 cm = 0.34 m

Magnetic field strength = B = 9000 mT = 9 T

Permeability of free space = μo = 4π × 10⁻⁷ T.m/A

Asked:

Number of turns = N = ?

Solution:

Grade: High School

Subject: Physics

Chapter: Magnetic Field

since you're a real machine, your ama is always less than your ima, because you can never totally eliminate friction.

when it gains or loses electrons