Numerical Problems on Current Carrying Solenoids

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Example – 01:

  • A solenoid has a core of material of relative permeability 4000. The number of turns is 1000 per metre. A current of 2 A flows through the solenoid. Find magnetic intensity, the magnetic field in core, magnetization, magnetic current and susceptibility.
  • Given: Relative permeability = μr = 4000, Number of turns per metre = n = 1000, Current flowing = i = 2A, μ= 4π x 10-7 Wb/Am.
  • To Find: magnetic intensity = H = ?, the magnetic field in core = B = ?, magnetization = M= ?, magnetic current = I=?, and susceptibility = χ = ?
  • Solution:

Magnetic intensity H = n i = 1000 x 2 = 2000 A/m

Intensity of magnetic field B = μ H = μr μ0 H = 4000 x 4π x 10-7 x 2000 = 10 T

Magnetization M= (μr  – 1)H = (4000 – 1) x 2000 = 3999 x 2000 = 7.998 x 10A/m



We have B = μr μ0 (i + Im)

10 =  4000 x 4π x 10-7 x (2 + Im)

(2 + Im) = 10 / (4000 x 4 x 3.142 x 10-7)

2 + Im = 1989



Magnetic current Im = 1987 A

We have μr = 1 + χ

∴  Susceptibility = χ = μ– 1 =  4000 – 1 = 3999

Example – 02:

  • A solenoid has 1000 turns and is 20 cm long. Find the magnetic induction produced at the centre of the solenoid by the current of 2 A. What is the flux at this point if the diameter of solenoid is 4 cm?
  • Given: Number of turns = N = 1000, Length  = l = 20 cm = 0.2 m, Current through solenoid = 2 A, diameter = 4 cm, radius of solenoid = 4/2 = 2 cm = 0.02 m, μ= 4π x 10-7 Wb/Am.
  • To Find: magnetic induction = B = ?, magnetic flux = Φ = ?
  • Solution:

n = N/= 1000/0.2 = 5000 turns per metre

B = μ H = μ n i = 4π x 10-7 x 5000 x 2 = 4 x 3.142 x 10-7 x 5000 x 2 = 0.01256 T



Now B = Φ/A

∴  Φ = B x A = B x πr2 =  0.01256 x 3.142 x (0.02)2

∴  Φ = 0.01256 x 3.142 x 4 x 10-4 =  1.58 x 10-5 Wb

Ans: Magnetic induction produced = 0.01256 T and magnetic flux = 1.58 x 10-5Wb

Example – 03:

  • A closely wound solenoid is 1 m long and has 5 layers of windings, each winding being of 500 turns. If the average diameter of the solenoid is 3 cm and it carries a current of 4 A, find the magnetic field at a point well within the solenoid.
  • Given: Number of turns = N =  500 x 5 = 2500, Length of solenoid = l = 1 m, Current through solenoid = 4 A, diameter = 3 cm, radius of solenoid = 3/2 = 1.5 cm = 0.015 m, μ= 4π x 10-7 Wb/Am.
  • To Find: magnetic induction = B = ?
  • Solution:

n = N/= 2500/1 = 2500 turns per metre



B = μ H = μ n i = 4π x 10-7 x 2500 x 4 = 4 x 3.142 x 10-7 x 2500 x 4 = 1.256 x 10-2 T

B = 0.01256 T

Ans: Magnetic induction produced = 0.01256 T

Example – 04:

  • A solenoid 0.5 m long has a four layer winding of 300 turns each. What current must pass through it to produce a magnetic field of induction 2.1 x 10-2 T at the centre.
  • Given: Number of turns = N =  300 x 4 = 1200, Length of solenoid = l = 0.5 m, Magnetic induction = B = 2.1 x 10-2 T,  μ= 4π x 10-7 Wb/Am.
  • To Find: Current through solenoid = i = ?
  • Solution:

n = N/= 1200/0.5 = 2400 turns per metre

 B = μ H = μ n i



∴  i = B/μ n = (2.1 x 10-2)/(4π x 10-7 x 2400) =  (2.1 x 10-2)/(4 x 3.142 x 10-7 x 2400)

∴  i = 6.96 A

Ans: Current through solenoid is 6.96 A.

Example – 05:

  • A solenoid (π/2) m long has a two-layer winding of 500 turns each and has radius 5 cm. What is the magnetic induction at the centre when it carries a current of 5 A.
  • Given: Number of turns = N =  500 x 2 = 1000, Length of solenoid = l = (π/2) m, Magnetic induction, Current through solenoid = i = 5 A.
  • To Find: Magnetic induction = B = ?
  • Solution:

n = N/= 1000/(π/2) = (2000/π) turns per metre



B = μ H = μ n i = 4π x 10-7 x (2000/π) x 5 = 4 x 10-3 T

Ans: Magnetic induction produced = 4 x 10-3 T

Example – 06:

  • A circular coil of 3000 turns per 0.6 m length carries a current of 1 A. What is the magnitude of magnetic induction (magnetic flux).
  • Given: Number of turns = n = 3000 per 0.6 m = 3000/0.6 = 5000 per metre, current through the coil = 1 A.
  • To Find: Magnetic induction = B = ?
  • Solution:

B = μ H = μ n i = 4π x 10-7 x 5000 x 1 = 4 x 3.142 x 10-7 x 5000 x 1 = 6.284 x 10-3 T

Ans: Magnetic induction produced = 6.284 x 10-3 T or 6.284 x 10-3 Wb/m2

Example – 07:

  • A solenoid of 20 turns per cm has a radius of 3 cm and is 40 cm long. Find the magnetic moment when it carries a current of 9.5 A.
  • Given: Number of turns = n = 20 per cm = 20/0.01 = 2000 per metre, radius of solenoid = 3 cm = 0.03 m, length of solenoid = l = 40 cm = 0.4 m, current through solenoid = 9.5 A.
  • To Find: Magnetic moment = M = ?
  • Solution:

Total turn of solenoid = N = n x l = 2000 x 0.4 = 800

M = N i A =  N i πr2 = 800 x 9.5 x 3.142 x (0.03)2 = 800 x 9.5 x 3.142 x 9 x 10-4 = 21.5 Am2



Ans: Magnetic moment is 21.5 Am2

Example – 08:

  • A circular coil of 300 turns and diameter 14 cm carries a current of 15 A. What is the magnitude of magnetic moment associated with the coil
  • Given: Number of turns = N = 300, diameter of coil = 14 cm, radius of coil = 14/2 = 7 cm = 0.07 m, current through the coil = 15 A.
  • To Find: Magnetic moment = M = ?
  • Solution:

M = N i A =  N i πr2 = 300 x 15 x 3.142 x (0.07)2 = 300 x 15 x 3.142 x 49 x 10-4 = 69.28 Am2

Ans: the magnetic moment is 69.28 Am2

Example – 09:

  • A closely wound solenoid of 1000 turns and area 4.2 cm2 carries a current of 3 A. It is suspended so as to move freely in horizontal plane in a horizontal magnetic field of  6 x 10-2 T. Find the magnetic moment, torque acting on the solenoid when the axis of solenoid makes an angle of 30° with the external horizontal field.
  • Given: Number of turns = N = 1000, Area of cross-section of solenoid = A = 4.2 cm2 = 4.2 x 10-4 m2, current through solenoid = 3 A, external magnetic field = B = 6 x 10-2 T, Angle made by solenoid axis with field = θ =30° ,
  • To Find: Magnetic moment = M = ?
  • Solution:

M = N i A =  1000 x 3 x 4.2 x 10-4 = 1.26 Am2



Now τ = MB sin θ = 1.26 x 6 x 10-2 x sin 30° =  1.26 x 6 x 10-2 x 0.5 = 0.0378 Nm

Ans: The magnetic moment is 1.26 Am and torque acting = 0.0378 Nm

Example – 10:

  • A closely wound solenoid of 1000 turns and area 2 x 10-4 mcarries a current of 1 A. It is placed in horizontal axis at 30° with the direction of uniform magnetic field of 0.16 T. Calculate the magnetic moment of solenoid and torque experienced by it in the field.
  • Given: Number of turns = N = 1000, Area of cross-section of solenoid = A = 2 x 10-4 m2, current through solenoid = 1 A, external magnetic field = B = 0.16 T, Angle made by solenoid axis with field = θ =30° ,
  • To Find: Magnetic moment = M = ?
  • Solution:

M = N i A =  1000 x 1 x 2 x 10-4 = 0.2 Am2

Now τ = MB sin θ = 0.2 x 0.16 x sin 30° =  0.2 x 0.16 x 0.5 = 0.016 Nm

Ans: The magnetic moment is 0.2 Am and torque acting = 0.016 Nm

Example – 11:

  • A closely wound solenoid of 2000 turns and area 1.6 x 10-4 mcarries a current of 4 A. It is in equilibrium with horizontal axis at 30° with the direction of uniform magnetic field of 7.5 x 10-2 T. Calculate the magnetic moment of the solenoid and also find the force and torque experienced by it in the field.
  • Given: Number of turns = N = 2000, Area of cross-section of solenoid = A = 1.6 x 10-4 m2, current through solenoid = 4 A, external magnetic field = B = 7.5 x 10-2  T, Angle made by solenoid axis with field = θ =30° ,
  • To Find: Magnetic moment = M = ?
  • Solution:

M = N i A =  2000 x 4 x 1.6 x 10-4 = 1.28 Am2



As the solenoid is in equilibrium position no force acts on it.

Now τ = MB sin θ = 1.28 x 7.5 x 10-2  x sin 30° = 1.28 x 7.5 x 10-2  x 0.5 = 0.048 Nm

Ans: The magnetic moment is 1.28 Am2, force = 0, and torque acting = 0.048 Nm

 

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