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Alecsey [184]
3 years ago
7

Two circular coils are concentric and lie in the same plane. The inner coil contains 140 turns of wire, has a radius of 0.015 m,

and carries a current of 7.2 A. The outer coil contains 180 turns and has a radius of 0.023 m. What must be the magnitude and direction (relative to the current in the inner coil) of the current in the outer coil, so that the net magnetic fi eld at the common center of the two coils is zero
Physics
1 answer:
Whitepunk [10]3 years ago
8 0

Given Information:  

Current = Ii = 7.2 A

Number of turns inner coil = Ni = 140

Number of turns outer coil = No = 180

Radius inner coil = ri = 0.015 m

Radius outer coil = ro = 0.023 m

Required Information:  

outer coil current = Io = ?

Answer:  

Io = 8.6 A

Step-by-step explanation:  

We have two coils, subscript i denotes inner coil and o denotes the outer coil.

According to the Biot-Savart Law  

B = μ₀IN/2r

we will equate the inner and outer magnetic fields

Bi=Bo

μ₀IiNi/2ri  = μ₀IoNo/2ro

μ₀ and 2 cancels out

Io = Ii*Ni*ro/No*ri

Io = 7.2*140*0.023/180*0.015

Io = 8.6 A

Therefore, a current of 8.6 A is needed to flow in opposite direction so that the magnetic field is opposite and the net magnetic field will be zero.

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Answer:

During start total turns

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Initially the machine is at rest and then starts rotating with speed 3550 rev/min

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now the number of turns is given as

N_4 = \frac{59.17 + 29.56}{2}(15.5)

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