Respuesta :
Answer:
(a)The current passes through the solenoid is 11.7 A.
(b) The new current will be one-fourth of the initial current.
Explanation:
Given that,
The number of turns per meter = 385
Diameter of solenoid = 17.0 cm = [tex]17\times 10^{-2}[/tex] m
Magnetic flux through core of solenoid [tex]\phi[/tex] = [tex]1.28\times 10^{-4}[/tex] Tm²
(a)
Magnetic field B= [tex]\mu_0nI[/tex]
[tex]\mu_0= 4\pi \times 10^{-7}[/tex] T/amp m
Cross section area of the solenoid A= [tex]\pi \frac{d^2}{4}[/tex]
[tex]=\pi\frac{ (17\times 10^{-2})^2}{4}[/tex] m²
The angle between magnetic field and cross section of the solenoid is [tex]\theta =0^\circ[/tex]
The magnetic flux through a area A with magnetic fie;d B is
[tex]\phi = BA cos\theta[/tex]
[tex]\Rightarrow \phi =( \mu_0nI)(\pi \frac{d^2}4)cos \theta[/tex]
[tex]\Rightarrow I =\frac{\phi}{(\mu_0\pi n \frac{d^2}4cos\theta)}[/tex]
[tex]=\frac{4\phi}{(\mu_0n)(\pi d^2)cos\theta}[/tex]
[tex]=\frac{1.28\times 10^{-4}\times 4}{(4\pi \times10^{-7}\times385 )\times(\pi\times17\times 10^{-2})^2cos 0^\circ}[/tex]
=11.7 A
The current of the solenoid is 11.7 A.
(b)
[tex]I =\frac{4\phi}{(\mu_0\pi n d^2cos\theta)}[/tex]
From the above equation it is clear that, the current is inversely proportional to the square of the diameter of a solenoid.
[tex]I\propto \frac1{d^2}[/tex].
Consider d' be the new diameter of the solenoid .
Since the new diameter of the solenoid is double of the initial diameter.
That is d'= 2d.
[tex]\frac{I}{I'}= \frac{(d')^2}{d^2}[/tex]
[tex]\Rightarrow \frac{I}{I'}=\frac{(2d)^2}{d^2}[/tex]
[tex]\Rightarrow \frac{I}{I'}=4[/tex]
⇒I=4I'
[tex]\Rightarrow I'=\frac{I}{4}[/tex]
The new current will be one-fourth of the initial current.
A) The current passing through the solenoid is = 11.7 Amperes
B) When the diameter of the solenoid is doubled the current = 2.93 A
Given data :
Number of turns ( n ) = 385
Diameter of solenoid ( d ) = 17.0 cm = 17 * 10⁻² m
Magnitude of magnetic flux ( ∅ ) = 1.28 * 10⁻⁴ T.m²
A) Determine the current in the solenoid
Current through the solenoid ( I ) = [ 4∅ / ( μο*n )*(πd²) *cosθ ) ] --- ( 1 )
= [ ( 4 * 1.28 * 10⁻⁴ ) / ( 4π * 10⁻⁷ * 385 ) * ( π * 17 * 10⁻² )² * cosθ ) ]
= 11.7 Amperes
B) when the diameter is doubled
( I ) = [ 4∅ / ( μο*n )*(πd²) *cosθ ) ] ----- ( 2 )
From the equation above
I ∝ 1/d² ( i.e. current in solenoid is inversely proportional to the diameter )
∴ when the diameter of the solenoid is doubled
The value of the new current ( I' ) = 1/4 * ( I )
= 1/4 * 11.7
≈ 2.93 amperes.
Hence we can conclude that The current passing through the solenoid is = 11.7 Amperes, When the diameter of the solenoid is doubled the current will be 2.93 A
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