Answer:
The current needed is 1790.26 A
Explanation:
Given;
magnitude of magnetic field, B = 1.5 T
length of the solenoid, L = 1.8 m
diameter of the solenoid, d = 75 cm = 0.75 m
The magnetic field is given by;
[tex]B = \frac{\mu_o NI }{L}[/tex]
Where;
μ₀ is permeability of free space = 4π x 10⁻⁷ m/A
I is current in the solenoid
N is the number of turns, calculated as;
[tex]N = \frac{Length \ of\ solenoid}{diameter \ of \ wire} \\\\N = \frac{1.8}{1.5*10^{-3}} =1200 \ turns[/tex]
The current needed is calculated as;
[tex]I = \frac{BL}{\mu_o N} \\\\I = \frac{1.5 *1.8}{4\pi *10^{-7} *1200} \\\\I = 1790.26 \ A[/tex]
Therefore, the current needed is 1790.26 A.
Answer:
I = 1790.5 A
Explanation:
The magnetic field due to a solenoid is given by the following formula:
B = μ₀NI/L
where,
B = Magnetic Field Required = 1.5 T
μ₀ = 4π x 10⁻⁷ T/A.m
L = length of Solenoid = 1.8 m
I = Current needed = ?
N = No. of turns = L/diameter of wire = 1.8 m/1.5 x 10⁻³ m = 1200
Therefore,
1.5 T = (4π x 10⁻⁷ T/A.m)(1200)(I)/1.8 m
I = (1.5 T)(1.8 m)/(1200)(4π x 10⁻⁷ T/A.m)
I = 1790.5 A
