A solenoid is designed to produce a magnetic field of 0.0270 T at its center. It

embaseclielenzn

embaseclielenzn

Answered question

2021-11-21

A solenoid is designed to produce a magnetic field of 0.0270 T at its center. It has radius 1.40 cm and length 40.0 cm, and the wire can carry a maximum current of 12.0 A. (a) What fewest number of turns per unit length must the solenoid have? (b) What entire length of wire is necessary?

Answer & Explanation

Lupe Kirkland

Lupe Kirkland

Beginner2021-11-22Added 21 answers

This problem concerns the magneteic field magnitude B at the center of acurrent-carrying solenoid, so we can use. We are given B=0.0270 T, I=12.0 A, r=1.40 cm, and L=40.0 cm. In part a) our target variable is n of the solenoid. In part b) the target variable is the total length l of the wire. We will assume the wire is tightly wound around the core of the solenoid.
Execute:
a) Solving equation for n and substituting the known values of B and I, we find
n=Bμ0I=0.0270 T(4π×107 TmA)(12.0 A)
=1790 turns/ m
b) Since the length of the solenoid is L=40.0 cm, the total number required for the solenoid is
N=nL=(1791 turnsm)(0.400 m)=716turns
Each turn is wound around the core of the solenoid as a circle of radius r=1.40 cm (so each turn amounts to 2πr length of the wire) and we have a total number of 716 turns, so the total length of the wire required is given by the product
l=N×2πr=(716 turns)(2π×0.014 m)=63.0m
Evaluate: The required length of the wire (the answer in part (b)) is appreciated. If the current supplied were even less, say 1 A, the required length of the wire would be 12 times larger than our answer, reaches 756 m.

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