You are presented with several wires made ofthe same conducting material. The radius and drift speed are givenfor each wire in terms of some

Question

You are presented with several wires made ofthe same conducting material. The radius and drift speed are givenfor each wire in terms of some unknown units \rm r and \rm v. Rank thewires in order of decreasing electron current.
Rank from most to least electroncurrent. To rank items as equivalent, overlap them.
radius=1 radius=4 radius =2 radius= 3
drift speed =5V drift speed = .5V drift speed = 2.5V drift speed = 1 V

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Khánh Gia 4 years 2021-08-16T02:06:07+00:00 2 Answers 9 views 0

Answers ( )

  1. trucchi
    0
    2021-08-16T02:07:13+00:00
    Reply

    Answer:

    So in order of most to least electric current,

    Ic > Id > Ib > Ia

    Explanation:

    Let the wires be represented as A, B, C, D

    in the the given order.

    So wire A has radius 1 and drift speed = 5V

    wire B has radius 4 and drift speed = 0.5V

    wire C has radius 2 and drift speed = 2.5V

    wire D has radius 3 nd drift speed = 1V

    The current in a given wire is directly proportional to the cross seectional area A(πr²) and also to the drift speed v

    I = nqvA

    Assuming n and q are constant,

    I ∝v× r²

    Ia = C×5v × r² = 5Cvr²

    Ib = C×0.5v × (4r)² = 8Cvr²

    Ic = C×2.5v× (2r)² = 10Cvr²

    Id = C×1v×(3r)² = 9Cvr²

    Where Cis a constant = nqπ

    So in order of most to least electric current,

    Ic > Id > Ib > Ia

  2. RuslanHeatt
    0
    2021-08-16T02:07:38+00:00
    Reply

    Answer:

    The ranking of current from the most to the least is I_{3} >I_{4}> I_{2}> I_{1}

    Explanation:

    Case Number  1 :

    Drift velocity=V_{drift}= 5v

    Radius=r= 1r

    Square of radius=r^{2}= 1

    Current=I=C(V _{drift} *r^{2} )=C(5vr^{2})

    Case Number 2 :

    Drift velocity=V_{drift}=0.5v

    Radius=r=4r

    Square of radius=r^{2}=16

    Current=I=C(V _{drift} *r^{2} )= C(8vr^{2})

    Case number 3 :

    Drift velocity=V_{drift}=2.5

    Radius=r=2r

    Square of radius=r^{2}=4

    Current=I=C(V _{drift} *r^{2} )= C(10vr^{2)

    Case number 4:

    Drift velocity=V_{drift}=1v

    Radius=r=3r

    Square of radius=r^{2}=9

    Current=I=C(V _{drift} *r^{2} )=C(9vr^{2)

    Hence, The ranking of current from the most to the least is I_{3} >I_{4}> I_{2}> I_{1}

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