- #1
predentalgirl1
- 67
- 1
1. A solenoid is 0.20 m long and consists of 100 turns of wire. At its center, the solenoid produces a magnetic field with a strength of 1.5 mT. Find the current in the coil.
2. L be the length of the solenoid = 0.2m
N be the number of turns = 100
B be the magnetic field = 1.5 mT = 1.5 x 10-3 T
I be the current in the coil in ampere
μ be the permeability in air = 4 π x 10-7 H/m
in a close circuit magnetic field
B = μNI / L
3. So, I = BL / μN
= (1.5x10-3 x 0.2) / (4 π x 10-7 x 100)
= 3 x 10-3 / 1256.637061 x 10-7
= (3/1256.637061) x 104
= 23.87324147 A
( Ans)
2. L be the length of the solenoid = 0.2m
N be the number of turns = 100
B be the magnetic field = 1.5 mT = 1.5 x 10-3 T
I be the current in the coil in ampere
μ be the permeability in air = 4 π x 10-7 H/m
in a close circuit magnetic field
B = μNI / L
3. So, I = BL / μN
= (1.5x10-3 x 0.2) / (4 π x 10-7 x 100)
= 3 x 10-3 / 1256.637061 x 10-7
= (3/1256.637061) x 104
= 23.87324147 A
( Ans)