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liubo4ka [24]
3 years ago
7

A long solenoid that has 1,140 turns uniformly distributed over a length of 0.415 m produces a magnetic field of magnitude 1.00

10-4 T at its center. What current is required in the windings for that to occur?
Physics
1 answer:
mezya [45]3 years ago
3 0

Answer:

Therefore,

Current required is , I

I = 0.0289\ Ampere

Explanation:

Given:

Turns = N = 1140

length of solenoid = l = 0.415 m

Magnetic Field,

B = 1.00\times 10^{-4}\ T

To Find:

Current , I = ?

Solution:

If N is the number of turns in the length, the total current through the rectangle is NI. Therefore, Ampere’s law applied to this path gives

\int {B} \, ds= Bl=\mu_{0}NI

Where,  

B = Strength of magnetic field

l = Length of solenoid

N = Number of turns

I = Current

\mu_{0}=Permeability\ in\ free\ space=4\pi\times 10^{-7}\ Tm/A

Therefore,

I =\dfrac{Bl}{\mu_{0}N}

Substituting the values we get

I =\dfrac{1.00\times 10^{-4}\times 0.415}{4\times 3.14\times 10^{-7}\times 1140}=0.0289\ Ampere

Therefore,

Current required is , I

I = 0.0289\ Ampere

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c. the volume of the part of the ship that lies below the water's surface.

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g Show that the first derivative of the magnitude of the net magnetic field of the coils (dB/dx) vanishes at the mid- point P re
never [62]

Answer:

Hi Carter,

The complete answer along with the explanation is shown below.

I hope it will clear your query

Pls rate me brainliest bro

Explanation:

The magnitude of the magnetic field on the axis of a circular loop, a distance z  from the loop center, is given by Eq.:

B = NμοiR² / 2(R²+Z²)³÷²

where

R is the radius of the loop

N is the number of turns

i is the current.

Both of the loops in the problem have the same radius, the same number of turns,  and carry the same current. The currents are in the same sense, and the fields they  produce are in the same direction in the region between them. We place the origin  at the center of the left-hand loop and let x be the coordinate of a point on the axis  between the loops. To calculate the field of the left-hand loop, we set z = x in the  equation above. The chosen point on the axis is a distance s – x from the center of  the right-hand loop. To calculate the field it produces, we put z = s – x in the  equation above. The total field at the point is therefore

B = NμοiR²/2 [1/ 2(R²+x²)³÷²   + 1/ 2(R²+x²-2sx+s²)³÷²]

Its derivative with respect to x is

dB /dx=  - NμοiR²/2 [3x/ (R²+x²)⁵÷²   + 3(x-s)/(R²+x²-2sx+s²)⁵÷² ]

When this is evaluated for x = s/2 (the midpoint between the loops) the result is

dB /dx=  - NμοiR²/2 [3(s/2)/ (R²+s²/4)⁵÷²   - 3(s/2)/(R²+s²/4)⁵÷² ] =0

independent of the value of s.

8 0
3 years ago
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