Question

The magnetic field strength at the north pole of a 2.0-cmcm-diameter, 8-cmcm-long Alnico magnet is 0.10 TT. To produce the same field with a solenoid of the same size, carrying a current of 2.5 AA , how many turns of wire would you need?

Answer 1

Answer:

The wire would need 2546 turns to produce the same magnetic field as solenoid

Explanation:

Given;

magnetic field strength, B = 0.1 T

diameter of solenoid, d = 2 cm, radius, r = 1 cm

length of solenoid, L = 8 cm

current in the wire, I = 2.5 A

Magnetic field inside solenoid = μ₀nI

Magnetic field for circular loop = (μ₀I / 2πr)

For equal magnetic field in solenoid and circular loop;

0.1 T = μ₀nI

$n = \frac{0.1}{4\pi *10^{-7}*2.5} = 31826.86 \ m^{-1}$

Number of loops, N = nL

N = 31826.86 m⁻¹ * 0.08 m = 2546 turns

Therefore, the wire would need 2546 turns to produce the same magnetic field as solenoid.

Answer 2

Given Information:  

Current = I = 2.5 A  

Magnetic field = B = 0.10 T  

Radius = r = d/2 = 0.02/2 = 0.01 m

Length = L = 8 cm = 0.08 m

Required Information:  

Number of turns = N = ?  

Answer:  

Number of turns = N ≈ 2547 turns

Step-by-step explanation:  

The approximate model to find the number of turns is given by

B = μ₀nI

Where n = N/L

so

B = μ₀NI/L

N = BL/μ₀I  

Where B is the magnetic field, L is the length of the solenoid, I is the current and μ₀ is the permeability of free space

N = (0.10*0.08)/(4πx10⁻⁷*2.5)

N ≈ 2547 Turns