The solenoid must have about 2332 turns
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Further explanation
Let's recall magnetic field strength from current carrying wire and from center of the solenoid as follows:
[tex]\boxed {B = \mu_o \frac{I}{2 \pi d} } [/tex]
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)
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[tex]\boxed {B = \mu_o \frac{I N}{L} } [/tex]
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 !
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Given:
Current = I = 4.3 A
Length = L = 42 cm = 0.42 m
Magnetic field strength = B = 0.030 T
Permeability of free space = μo = 4π × 10⁻⁷ T.m/A
Asked:
Number of turns = N = ?
Solution:
[tex]B = \mu_o \frac{I N}{L}}[/tex]
[tex]\frac{I N}{L} = B \div \mu_o[/tex]
[tex]IN = BL \div \mu_o[/tex]
[tex]N = BL \div (\mu_o I)[/tex]
[tex]N = ( 0.030 \times 0.42 ) \div ( 4 \pi \times 10^{-7} \times 4.3 )[/tex]
[tex]\boxed {N \approx 2332}[/tex]
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Learn more
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Answer details
Grade: High School
Subject: Physics
Chapter: Magnetic Field
