Question

If coil radius = 20 cm, length of coil 100 cm, magnetic field constant $1.257*10^{-6}$, relatively permeability=1 and the inductance, the inductance=1.58mH , the number of turns in wire coil is O a. 1 O b. 100 O c. 10 O d. 1000 Length (l) Inner Core Material Cross-sectional Area. (A) Number of Turns (N)

Name: if coil radius 20 cm length of coil 100 cm magnetic field constant 125710 6 relatively permeability1 and the inductance the inductance158mh the number of turns in wire coil is o a 1 o b 100 35182
Uploaded: 2022-05-26T12:43:30-08:00
Duration: 2 min 54 s
Channel: Adi S
Description: if coil radius 20 cm length of coil 100 cm magnetic field constant 125710 6 relatively permeability1 and the inductance the inductance158mh the number of turns in wire coil is o a 1 o b 100 35182

          If coil radius = 20 cm, length of coil 100 cm, magnetic field constant $1.257*10^{-6}$, relatively permeability=1 and the inductance, the inductance=1.58mH , the number of turns in wire coil is
O a. 1
O b. 100
O c. 10
O d. 1000
Length (l)
Inner Core
Material
Cross-sectional
Area. (A)
Number of Turns (N)

if coil radius 20 cm length of coil 100 cm magnetic field constant 125710 6 relatively permeability1 and the inductance the inductance158mh the number of turns in wire coil is o a 1 o b 100 35182

Added by Sergio B.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Adi S

05/26/2022

Step 1

Given: Coil radius, $r = 20 \text{ cm} = 0.20 \text{ m}$ Length of coil, $l = 100 \text{ cm} = 1.00 \text{ m}$ Magnetic field constant (permeability of free space), $\mu_0 = 1.257 \times 10^{-6} \text{ H/m}$ (Note: The text says "magnetic field constant Show more…

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If coil radius = 20 cm, length of coil 100 cm, magnetic field constant $1.257*10^{-6}$, relatively permeability=1 and the inductance, the inductance=1.58mH , the number of turns in wire coil is O a. 1 O b. 100 O c. 10 O d. 1000 Length (l) Inner Core Material Cross-sectional Area. (A) Number of Turns (N)