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Papessa [141]
3 years ago
6

The magnetic field strength at the north pole of a 2.0-cmcm-diameter, 8-cmcm-long Alnico magnet is 0.10 TT. To produce the same

field with a solenoid of the same size, carrying a current of 2.4 AA , how many turns of wire would you need
Physics
1 answer:
horrorfan [7]3 years ago
6 0

Answer:

2653 turns

Explanation:

We are given that

Diameter,d=2 cm

Length of magnet,l=8 cm=8\times 10^{-2} m

1m=100 cm

Magnetic field,B=0.1 T

Current,I=2.4 A

We are given that

Magnetic field of solenoid and magnetic are same and size of both solenoid and magnetic are also same.

Length of solenoid=8\times 10^{-2} m

Magnetic field of solenoid

B=\frac{\mu_0NI}{l}

Using the formula

0.1=\frac{4\pi\times 10^{-7}\times 2.4\times N}{8\times 10^{-2}}

Where \mu_0=4\pi\times 10^{-7}

N=\frac{0.1\times 8\times 10^{-2}}{4\pi\times 10^{-7}\times 2.4}=2653 turns

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