Answer:
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
[tex]B = \frac{\mu _{o}N i}{L}[/tex]
[tex]9\times 10^{-3} = \frac{(4\pi\times 10^{-7} )N (2)}{0.34}[/tex]
N = 1218
The value of N is about B) 1218
[tex]\texttt{ }[/tex]
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)
[tex]\texttt{ }[/tex]
[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 !
[tex]\texttt{ }[/tex]
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:
[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 = ( 9 \times 0.34 ) \div ( 4 \pi \times 10^{-7} \times 2000 )[/tex]
[tex]\boxed {N \approx 1218}[/tex]
[tex]\texttt{ }[/tex]
[tex]\texttt{ }[/tex]
Grade: High School
Subject: Physics
Chapter: Magnetic Field
