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Chapter 29: Q52P (page 861)

A solenoid 1.30 long and2.60cm in diameter carries a current of 1.80A. The magnetic field inside the solenoid is 23.0mT . Find the length of the wire forming the solenoid.

Short Answer

Expert verified
  1. The length of the wire forming the solenoid is 108m.

Step by step solution

01

Listing the given quantities 

l=1.30m

d=2.60cm=0.0260m

B=23mT=0.023T

i=18A

02

Understanding the concept of magnetic field and solenoid

We find the number of turns of the solenoid. Using the number of turns, we can find the total length of the wire used in making the solenoid.

B=μ0in=μ0iNl

03

Calculations of the length of the wire forming the solenoid  

The number of turns of the solenoid is

B=μ0iNl

Nl=Bμ0i

N=Blμ0i=0.023×1.301.26×10−6×18≈1322

As we know that the length of the circular loop is the circumference of that loop. So, the total length of the solenoid is

L=2πrN

Since r=d2

Thus,

L=2π×0.02602×1322=π×0.0260×1322=107.98≈108m

The length of the wire forming the solenoid is108m .

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Most popular questions from this chapter

Figure 29-81 shows a wire segment of length Δs=3cm, centered at the origin, carrying current i=2A in the positive ydirection (as part of some complete circuit). To calculate the magnitude of the magnetic field produced by the segment at a point several meters from the origin, we can use B→=μ04πiΔs→×r^r2 as the Biot–Savart law. This is because r and u are essentially constant over the segment. Calculate (in unit-vector notation) at the(x,y,z)coordinates (a)localid="1663057128028" (0,0,5m)(b)localid="1663057196663" (0,6m,0)(c) localid="1663057223833" (7m,7m,0)and (d)(-3m,-4m,0)

Figure 29-25 represents a snapshot of the velocity vectors of four electrons near a wire carrying current i. The four velocities have the same magnitude; velocity is directed into the page. Electrons 1 and 2 are at the same distance from the wire, as are electrons 3 and 4. Rank the electrons according to the magnitudes of the magnetic forces on them due to current i, greatest first.

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