MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
Grade: 11
        Assume that each iron atom has a permanent magnetic moment equal to 2 Bohr magnetons (1 Bohr magnetic equals 9.27 × 10^-24 A-m^2). The density of atoms in iron is 8.52 × 10^28 atoms/m^3. (a) Find the maximum magnetization I in a long cylinder of iron. (b) Find the maximum magnetic field B on the axis inside the cylinder.
5 years ago

Answers : (1)

Aditi Chauhan
askIITians Faculty
396 Points
							Sol. 
f = 8.52 x 10 ^28 atoms / m^3
For maximum ‘I’, Let us consider the no. of atoms present in 1 m^3 of volume.
Given: m per atom = 2 × 9.27 × 10^–24 A–m^2
I = net m / V = 2 × 9.27 × 10^–24 × 8.52 × 10^28 ≈ 1.58 × 10^6 A/m
B = μ0 (H + I) = μ0  [∴ H = 0 in this case]
= 4π × 10^–7 × 1.58 × 10^6 = 1.98 × 10^–1 ≈ 2.0 T

						
5 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

  • 731 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution


Course Features

  • 57 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details