Home
Class 12
PHYSICS
A solenoid has 10^(3) turns per unit len...

A solenoid has `10^(3)` turns per unit length. On passing a current of 2A, magnetic induction is measured to be `4 pi (Wb) / (m)^(2)`. Calculate the magnetic susceptibility of the core.

Text Solution

AI Generated Solution

The correct Answer is:
To calculate the magnetic susceptibility of the core of the solenoid, we can follow these steps: ### Step 1: Understand the relationship between magnetic induction (B), magnetic field strength (H), and magnetic permeability (μ). The magnetic induction (B) in a solenoid can be expressed as: \[ B = \mu H \] where \( \mu \) is the permeability of the material and \( H \) is the magnetic field strength. ### Step 2: Calculate the magnetic field strength (H). The magnetic field strength (H) in a solenoid is given by: \[ H = N \cdot I \] where \( N \) is the number of turns per unit length, and \( I \) is the current flowing through the solenoid. Given: - \( N = 10^3 \, \text{turns/m} \) - \( I = 2 \, \text{A} \) Substituting the values: \[ H = 10^3 \cdot 2 = 2000 \, \text{A/m} \] ### Step 3: Use the given value of magnetic induction (B) to find the permeability (μ). We know from the problem statement that: \[ B = 4\pi \, \text{Wb/m}^2 \] Using the relationship \( B = \mu H \): \[ \mu = \frac{B}{H} \] Substituting the values we have: \[ \mu = \frac{4\pi}{2000} \] ### Step 4: Calculate the value of μ. Using \( \pi \approx 3.14 \): \[ \mu \approx \frac{4 \times 3.14}{2000} = \frac{12.56}{2000} = 0.00628 \, \text{Wb/(A·m)} \] ### Step 5: Relate permeability (μ) to the relative permeability (μr). The permeability of free space (μ₀) is approximately: \[ \mu_0 \approx 4\pi \times 10^{-7} \, \text{Wb/(A·m)} \] The relative permeability (μr) is given by: \[ \mu_r = \frac{\mu}{\mu_0} \] Substituting the values: \[ \mu_r = \frac{0.00628}{4\pi \times 10^{-7}} \] ### Step 6: Calculate μr. Using \( \pi \approx 3.14 \): \[ \mu_r \approx \frac{0.00628}{4 \times 3.14 \times 10^{-7}} \] \[ \mu_r \approx \frac{0.00628}{1.256 \times 10^{-6}} \approx 5000 \] ### Step 7: Calculate the magnetic susceptibility (χ). The magnetic susceptibility (χ) is related to the relative permeability (μr) as follows: \[ \chi = \mu_r - 1 \] Substituting the value of μr: \[ \chi = 5000 - 1 = 4999 \] ### Final Answer: The magnetic susceptibility of the core is: \[ \chi = 4999 \] ---
Promotional Banner

Topper's Solved these Questions

  • MAGNETIC FIELD AND MAGNETIC FORCES

    CENGAGE PHYSICS|Exercise Multiple Correct Answer type|2 Videos
  • MECHANICAL PROPERTIES OF FLUIDS

    CENGAGE PHYSICS|Exercise Question Bank|5 Videos

Similar Questions

Explore conceptually related problems

A long solenoid has 200 turns per cm and carries a current of 2.5A. The magnetic field at its centre is

The magnetic field (B) inside a long solenoid having n turns per unit length and carrying current / when iron core is kept in it is (mu_o = permeability of vacuum, chi = magnetic susceptibility)

A solenoid having 500 turns and length 2 m has radius of 2 cm , the self inductance of solenoid

A long solenoid has 800 turns per metre length of solenoid A current of 1.6 A flow through it. The magnetic induction at the end of the solenoid on its axis is

There is one current carrying solenoid with 750 turns per metre. A current of 2 A is flowing through the solenoid. A rod of certain material is inserted as core inside this solenoid and the intensity of magnetization is found to be 0.12 A/m. What is magnetic susceptibility of the material used? What kind of material is this?

A long solenoid with 40 turns per cm carries a current of 1 A. The magnetic energy stored per unit volume is..... J m^(-3) .

CENGAGE PHYSICS-Magnetism and Matter-Question Bank
  1. Relation between permeability' mu and magnetizing field H for a sample...

    Text Solution

    |

  2. When arod of magnetic material of size 10cm xx 0.5 cm xx 0,2cm is loc...

    Text Solution

    |

  3. A solenoid has 10^(3) turns per unit length. On passing a current of 2...

    Text Solution

    |

  4. A bar magnet of magnetic moment M and moment of inertia I (about cente...

    Text Solution

    |

  5. A magnet makes 40 oscillations per minute at a place having magnetic f...

    Text Solution

    |

  6. A puramagnetic sample shows a net magnetization of '8 A m^(-1) when pl...

    Text Solution

    |

  7. A solenoid of 500 turns /m is carrying a current of 3A. Relative perm...

    Text Solution

    |

  8. A rod of magnetic material of cross section 0.25 cm^(2) is located in ...

    Text Solution

    |

  9. Magnetic field of the earth is 0.3 G. A magnet is oscillating with the...

    Text Solution

    |

  10. A magnetic dipole is under the influence of two magnetic fields. The a...

    Text Solution

    |

  11. The area of hysteresis loop. of a material is equivalent to 250J / m^2...

    Text Solution

    |

  12. Density of iron is 7.8 g/ (cc) and induced 'nagnetic field in iron is1...

    Text Solution

    |

  13. A closely wound solenoid of 3000 turns and area of cross-section 2 xx ...

    Text Solution

    |

  14. At a given place on the earth's surface, the horizontal component of e...

    Text Solution

    |

  15. At a certain place, the horizontal component of the earth's magnetic f...

    Text Solution

    |

  16. At a certain locatioa in Africa, compass point 12^(circ) west of geogr...

    Text Solution

    |

  17. A dipole of magnetic moment vec(m)=(30 hati) (A) (m)^(2) is placed alo...

    Text Solution

    |

  18. A vibration magnetometer placed in magnetic meridian has a small bar m...

    Text Solution

    |

  19. A bar magnet is hung by a thin cotton thread in a uniform horizontal m...

    Text Solution

    |

  20. The magnetic moment of a magnet of mass 150 g is 18 xx 10^(-3) A m^(2)...

    Text Solution

    |