# 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?

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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 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=μSolution 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.

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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.

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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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Guys, does anyone know the answer?