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**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

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)

= (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)