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Magnetic field lines can be entirely confined within the core of a toroid, but not within a straight solenoid Why? - Physics

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प्रश्न

Magnetic field lines can be entirely confined within the core of a toroid, but not within a straight solenoid Why?

टीपा लिहा
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उत्तर

  1. The magnetic field around the toroid consists of concentric circular lines of force around it. Thus, the magnetic field in the interior of the toroid is tangential to each loop.
  2. Whereas, the magnetic field produced by a solenoid is similar to the magnetic field of a bar magnet. One end of the solenoid coil acts as a south pole and the other end acts as the north pole with the field lines inside the solenoid remaining parallel. Thus, the magnetic field B is parallel to the axis of the solenoid.

As a result, magnetic field lines can be entirely confined within the core of a toroid, but not within a straight solenoid.  

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Applications of Ampere’s Circuital Law > Magnetic Field of a Toroidal Solenoid
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पाठ 10: Magnetic Effect of Electric Current - Short Answer I

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एससीईआरटी महाराष्ट्र Physics [English] 12 Standard HSC
पाठ 10 Magnetic Effect of Electric Current
Short Answer I | Q 2

संबंधित प्रश्‍न

A toroid of a central radius of 10 cm has windings of 1000 turns. For a magnetic field of 5 × 10-2 T along its central axis, what current is required to be passed through its windings?


What is Solenoid?


What is Toroid?


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


Using Ampere's Law, derive an expression for the magnetic induction inside an ideal solenoid carrying a steady current.


A winding wire which is used to frame a solenoid can bear a maximum of 20 A current. If the length of the solenoid is 80 cm and its cross-sectional radius is 3 cm, then the required length of winding wire is ______ (B = 0.2 T)


The ratio of magnetic field and magnetic moment at the centre of a current carrying circular loop is x. When both the current and radius is doubled then the ratio will be ______.


Two current-carrying coils have radii r and 4r and have same magnetic induction at their centres. The ratio of voltage applied across them is ______.


The magnetic induction along the axis of a toroidal solenoid is independent of ______.


A proton is projected with a uniform velocity 'v' along the axis of a current carrying solenoid, then ____________.


The space within a current carrying toroid is filled with a m metal of susceptibility 16.5 x 10-6. The percentage increase in the magnetic field B is ____________.


Magnetic induction due to a toroid does not depend upon ______.


A toroid is a long coil of wire wound over a circular core. If 'r' and 'R' are the radii of the coil and toroid respectively, the coefficient of self-induction of the toroid is (The magnetic field in it is uniform and R > > r) ____________.

(N = number of turns of the coil and µ0 = permeability of free space)


A long solenoid carrying current 'I1' produces magnetic field 'B1' along its axis. If the current is reduced to 25% and number of turns per cm are increased four times, then new magnetic field 'B2' is ____________.


A winding wire is used to prepare a solenoid that can bear a maximum current of 10A. If the length of a solenoid is 80 cm and its cross-sectional radius is 3 cm, the required length of winding wire is ____________.

(magnetic field B = 0.2 T, µ0 = 4 x 10-7 SI units)


A charged particle carrying a charge 'q' and moving with velocity V, enters into a solenoid carrying a current T, along its axis. If 'B' is the magnetic induction along the axis of a solenoid, then the force 'F' acting on the charged particle will be ____________.


In a current-carrying long solenoid, the field produced does not depend upon ______


A long solenoid has 200 turns per cm and carries a current of 2.5 A. The magnetic field at the center is ______. (µ0 = 4π × 10-7 Wb/m-A)


A current of 10 A passes through a coil having 5 turns and produces magnetic field at the centre of the coil having magnitude 0.5 x 10-4 T. Calculate diameter of the coil.

0 = 4π x 10-7 Wb/Am)  


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