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4 years ago

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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?

1 answer:

mezya [45]4 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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Solving for t, we find

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And added to the 3 seconds between the instant of the jump and the moment he opens the parachute, the total time is

t = 3 s + 2.24 s = 5.24 s

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