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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 = μ_{r}= 4000, Number of turns per metre = n = 1000, Current flowing = i = 2A, μ_{o }= 4π x 10^{-7}Wb/Am.**To**Find:_{Z }= ?, magnetic current = I_{m }=?, and susceptibility = χ = ?**Solution:**

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

Intensity of magnetic field B = μ H = μ_{r} μ_{0} H = 4000 x 4π x 10^{-7 }x 2000 = 10 T

Magnetization M_{Z }= (μ_{r} – 1)H = (4000 – 1) x 2000 = 3999 x 2000 = 7.998 x 10^{6 }A/m

We have B = μ_{r} μ_{0} (i + I_{m})

10 = 4000 x 4π x 10^{-7 }x (2 + I_{m})

(2 + I_{m}) = 10 / (4000 x 4 x 3.142 x 10^{-7})

2 + I_{m} = 1989

Magnetic current I_{m} = 1987 A

We have μ_{r} = 1 + χ

∴ Susceptibility = χ = μ_{r }– 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_{o }= 4π x 10^{-7}Wb/Am.**To Find:**magnetic induction = B = ?, magnetic flux = Φ = ?**Solution:**

n = N/*l *= 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 πr^{2} = 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^{-5}Wb

#### 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_{o }= 4π x 10^{-7}Wb/Am.**To Find:**magnetic induction = B = ?**Solution:**

n = N/*l *= 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_{o }= 4π x 10^{-7}Wb/Am.**To Find:**Current through solenoid = i = ?**Solution:**

n = N/*l *= 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/*l *= 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/m^{2}

#### 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 πr^{2} = 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 Am^{2}

**Ans:** Magnetic moment is 21.5 Am^{2}

#### 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 πr^{2} = 300 x 15 x 3.142 x (0.07)^{2} = 300 x 15 x 3.142 x 49 x 10^{-4} = 69.28 Am^{2}

**Ans:** the magnetic moment is 69.28 Am^{2}

#### Example – 09:

- A closely wound solenoid of 1000 turns and area 4.2 cm
^{2}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 cm^{2}= 4.2 x 10^{-4}m^{2}, 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 Am^{2}

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^{2 } and torque acting = 0.0378 Nm

#### Example – 10:

- A closely wound solenoid of 1000 turns and area 2 x 10
^{-4}m^{2 }carries 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}m^{2}, 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 Am^{2}

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^{2 } and torque acting = 0.016 Nm

#### Example – 11:

- A closely wound solenoid of 2000 turns and area 1.6 x 10
^{-4}m^{2 }carries 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}m^{2}, 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 Am^{2}

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 Am^{2}, force = 0, and torque acting = 0.048 Nm

Science > Physics > Magnetism >You are Here |

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