As a new electrical technician, you are designing a large solenoid to produce a uniform 0.160-t magnetic field near the center of the solenoid. you have enough wire for 4300 circular turns. this solenoid must be 1.10 m long and 3.50 cm in diameter. what current will you need to produce the necessary field?

Current = 32.6 A

Explanation:

Magnetic field = B = 0.160 T

No. of turns of coil = N = 4300

Length of the solenoid = L = 1.10 m

Current required depends on the B, L , N and the magnetic permeability μ₀

B =  μ₀ N I / L

⇒ Current =  I = B L /  μ₀N = ( 0.160 ) ( 1.10) / (4π × 10⁻⁷)(4300)

                                           =  32.6 A

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