A solenoid 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 gen

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4 years ago

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A solenoid 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 gen

erated 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

Physics

2 answers:

sweet [91] · Answer 1 · 2021-11-16T06:46:16+03:00

The value of N is about B) 1218

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<h3>Further explanation</h3>

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

$\boxed {B = \mu_o \frac{I}{2 \pi d} }$

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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$\boxed {B = \mu_o \frac{I N}{L} }$

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 = 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:

$B = \mu_o \frac{I N}{L}}$

$\frac{I N}{L} = B \div \mu_o$

$IN = BL \div \mu_o$

$N = BL \div (\mu_o I)$

$N = ( 9 \times 0.34 ) \div ( 4 \pi \times 10^{-7} \times 2000 )$

$\boxed {N \approx 1218}$

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<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Magnetic Field

umka21 [38] · Answer 2 · 2021-11-16T05:45:18+03:00

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

$B = \frac{\mu _{o}N i}{L}$

$9\times 10^{-3} = \frac{(4\pi\times 10^{-7} )N (2)}{0.34}$

N = 1218