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    a long solenoid of length 1 m and having 1000 turn carries a current of 5 a. the magnetic field near one end of the solenoid is

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    A solenoid of length 1.0 m , radius 1 cm and total turns 1000 wound on it, carries a current of 5 A . Calculate the magnitude of the axial magnetic field inside the solenoid. If an element was to move with a speed of 104 m/s along the axis of this current carrying solenoid, what would be the force experienced by this electron?

    Click here👆to get an answer to your question ✍️ A solenoid of length 1.0 m , radius 1 cm and total turns 1000 wound on it, carries a current of 5 A . Calculate the magnitude of the axial magnetic field inside the solenoid. If an element was to move with a speed of 104 m/s along the axis of this current carrying solenoid, what would be the force experienced by this electron?

    Question

    A solenoid of length 1.0 m, radius 1 cm and total turns 1000 wound on it, carries a current of 5 A. Calculate the magnitude of the axial magnetic field inside the solenoid. If an element was to move with a speed of 104m/s along the axis of this current carrying solenoid, what would be the force experienced by this electron?

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    Updated on : 2022-09-05

    B=μ

    Solution Verified by Toppr 0 ​ ni = μ 0 ​ l n ​ i = (4π×10 −7 ) 1 1000 ​ ×5 B=2π×10 −3

    force on the electron= qVB

    = (1.6×10 −19 )×104×2π×10 −3 N Fe=1.044×10 −19 N

    Hence the force experienced by this electron is Fe=1.044×10

    −19 N

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    A solenoid, of length 1.0 m, has a radius of 1 cm and has a total of 1000 turns wound on it. It carries a current of 5 A. Calculate the magnitude of the axial magnetic field inside the solenoid. If an electron were to move with a speed of 10^(4) m s^(

    Here length of solenoid l = 1.0m, total no. of turns N = 1000 and current flowing l = 5 A :. Magnetic field inside the solenoid along its axial line B = mu0 N/l cdot I = 4 pi xx 10^(-7) xx 1000/1.0 xx 5 = 6.28 xx 10^(-3)T. As electron is to move with a speed a 10^(4) m s^(-1) along the axis of solenoid , hence vecv and vecB are along the same line. Hence, force experienced with this eletron will be zero.

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    A solenoid, of length 1.0 m, h...

    A solenoid, of length 1.0 m, has a radius of 1 cm and has a total of 1000 turns wound on it. It carries a current of 5 A. Calculate the magnitude of the axial magnetic field inside the solenoid. If an electron were to move with a speed of

    10 4 m s −1 104ms-1

    along the axis of this current carrying solenoid, what would be the force experienced by electron ?

    Text Solution Open Answer in App Solution

    Here length of solenoid l = 1.0m, total no. of turns

    N=1000 N=1000

    and current flowing l = 5 A

    ∴ ∴

    Magnetic field inside the solenoid along its axial line

    B= μ 0 N l ⋅I=4π× 10 −7 × 1000 1.0 ×5=6.28× 10 −3 T

    B=μ0Nl⋅I=4π×10-7×10001.0×5=6.28×10-3T

    .

    As electron is to move with a speed a

    10 4 m s −1 104ms-1

    along the axis of solenoid , hence

    v ⃗ v→ and B ⃗ B→

    are along the same line. Hence, force experienced with this eletron will be zero.

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    A solenoid of length π m and 5 cm in diameter has winding of 1000 turns and carries a current of 5 A. Calculate the magnetic field at its center along the axis.

    A solenoid of length π m and 5 cm in diameter has winding of 1000 turns and carries a current of 5 A. Calculate the magnetic field at its center along the axis.

    Advertisement Remove all ads Numerical

    A solenoid of length π m and 5 cm in diameter has winding of 1000 turns and carries a current of 5 A. Calculate the magnetic field at its center along the axis.

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    SOLUTION

    Data: L= 3.142 m, N = 1000, l = 5A,

    μ0 = 4π × 10-7 T m/A

    The magnetic induction,

    B = nI N L I μ0nI=μ0(NL)I

    =(4π×10-7)(10003.142)(5)=20×3.142×10-43.142

    = 2 × 10-3 T

    Concept: Magnetic Field of a Solenoid and a Toroid

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    Chapter 10: Magnetic Fields due to Electric Current - Exercises [Page 250]

    Q 15 Q 14 Q 16

    APPEARS IN

    Balbharati Physics 12th Standard HSC Maharashtra State Board

    Chapter 10 Magnetic Fields due to Electric Current

    Exercises | Q 15 | Page 250

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