Hard magnetic materials are difficult to magnetize and demagnetize, such as tungsten steel, ferrite, neodymium-iron-boron, etc. These materials are easy to magnetize and difficult to demagnetize. They are mainly used as tape materials and permanent magnets. Soft magnetic materials refer to ferromagnetic materials with narrow hysteresis loops, high permeability, low remanence, low coercivity and low magneto resistance. Difference Between Soft and Hard Magnetic Materials are as Follows:
5.Soft iron acquires saturation magnetisation for quite low value of magnetising field than in case of steel or soft iron is much strongly magnetised than steel.
Ex. An iron sample of mass 8.4 kg is repeatedly magnetised and demagnetised at a frequency of 50 cycles/sec. $3.2 \times 10^{4}$ Joule of energy is lost as heat in 30 minutes. If density of iron is 7200 $\mathrm{kg} / \mathrm{m}^{3}$ calculate the value of energy dissipated per unit volume per cycle in iron sample. Sol. If H is energy lost per unit volume per second in iron sample then energy lost by sample in a given time t is $E=H V n t$ where V is volume of sample So $\quad \mathrm{H}=\frac{\mathrm{E}}{\mathrm{Vnt}}=\frac{3.2 \times 10^{4} \times 7200}{8.4 \times 50 \times 30 \times 60}=304.8 \mathrm{Jm}^{-3} \mathrm{cycle}^{-1}$ dia, para and ferromagnetic substances

Comparison of properties of Soft iron and Steel
- The area of hysteresis loop for soft iron is much smaller than for steel so energy loss per unit volume per cycle or soft iron is smaller than steel.
- The retentivity of soft iron is greater than that of steel.
- The coercivity of steel is much larger than that of soft iron.
- The magnetisation and demagnetisation is easier in soft iron than steel.

Ex. An iron sample of mass 8.4 kg is repeatedly magnetised and demagnetised at a frequency of 50 cycles/sec. $3.2 \times 10^{4}$ Joule of energy is lost as heat in 30 minutes. If density of iron is 7200 $\mathrm{kg} / \mathrm{m}^{3}$ calculate the value of energy dissipated per unit volume per cycle in iron sample. Sol. If H is energy lost per unit volume per second in iron sample then energy lost by sample in a given time t is $E=H V n t$ where V is volume of sample So $\quad \mathrm{H}=\frac{\mathrm{E}}{\mathrm{Vnt}}=\frac{3.2 \times 10^{4} \times 7200}{8.4 \times 50 \times 30 \times 60}=304.8 \mathrm{Jm}^{-3} \mathrm{cycle}^{-1}$ dia, para and ferromagnetic substances
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All Study Material
- JEE Main
- Exam Pattern
- Previous Year Papers
- PYQ Chapterwise
- Physics
- Kinematics 1D
- Kinemetics 2D
- Friction
- Work, Power, Energy
- Centre of Mass and Collision
- Rotational Dynamics
- Gravitation
- Calorimetry
- Elasticity
- Thermal Expansion
- Heat Transfer
- Kinetic Theory of Gases
- Thermodynamics
- Simple Harmonic Motion
- Wave on String
- Sound waves
- Fluid Mechanics
- Electrostatics
- Current Electricity
- Capacitor
- Magnetism and Matter
- Electromagnetic Induction
- Atomic Structure
- Dual Nature of Matter
- Nuclear Physics
- Radioactivity
- Semiconductors
- Communication System
- Error in Measurement & instruments
- Alternating Current
- Electromagnetic Waves
- Wave Optics
- X-Rays
- All Subjects
- Physics
- Motion in a Plane
- Law of Motion
- Work, Energy and Power
- Systems of Particles and Rotational Motion
- Gravitation
- Mechanical Properties of Solids
- Mechanical Properties of Fluids
- Thermal Properties of matter
- Thermodynamics
- Kinetic Theory
- Oscillations
- Waves
- Electric Charge and Fields
- Electrostatic Potential and Capacitance
- Current Electricity
- Thermoelectric Effects of Electric Current
- Heating Effects of Electric Current
- Moving Charges and Magnetism
- Magnetism and Matter
- Electromagnetic Induction
- Alternating Current
- Electromagnetic Wave
- Ray Optics and Optical Instruments
- Wave Optics
- Dual Nature of Radiation and Matter
- Atoms
- Nuclei
- Semiconductor Electronics: Materials, Devices and Simple Circuits.
- Chemical Effects of Electric Current,
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