An iron bar of length 10 cm and diameter 2 cm is placed in a magnetic field of intensity 1000 Am^(-1) with its length….

Q: An iron bar of length 10 cm and diameter 2 cm is placed in a magnetic field of intensity 1000 Am-1 with its length parallel to the direction of the field. Determine the magnetic moment produced in the bar if permeability of its material is 6.3 × 10-4 T mA-1

Sol: We know that, μ = μ0 (1 + χ)

χ = μ/μ0 – 1

χ = (6.3 × 10-4)/(4π × 10-7 ) – 1 = 500.6

Intensity of magnetisation,

I = χ H = 500.6 × 1000 = 5 × 105 Am-1

∴ Magnetic moment, M = I × V = I × π r2 l

= 5 × 105 × 3.14 × (10-2)2  × (10 × 10-2) = 17.70 A-m2

A magnetic field strength (H) 3 × 10^3 Am^(-1) produce a magnetic field of induction (B) of 12π T in an iron rod…

Q: A magnetic field strength (H) 3 × 103 Am-1 produce a magnetic field of induction (B) of 12π T in an iron rod. Find the relative permeability of iron ?

Sol: $\large \mu = \frac{B}{H} = \frac{12 \pi}{3 \times 10^3}$

= 4π × 10-3

$\large \mu_r = \frac{\mu}{\mu_0} = \frac{4\pi \times 10^{-3}}{4\pi \times 10^{-7}}$

= 104

The magnetic moment of magnet of mass 75 gm is 9 × 10^(-7) A-m^2. If the density of the material of magnet is…

Q: The magnetic moment of magnet of mass 75 gm is 9 × 10-7 A-m2 . If the density of the material of magnet is 7.5 × 103 kg m-3 , then find intensity of magnetisation .

Sol: I = M/V Where Volume, V = (mass(m))/(density(p))

$\large I = \frac{M \times \rho}{m}$

$\large I = \frac{9 \times 10^{-7} \times 7.5 \times 10^3}{75 \times 10^{-3}}$

= 0.09 A/m.

The permeability of substance is 6.28 × 10^(-4) wb/A-m. Find its relative permeability and susceptibility ?

Q: The permeability of substance is 6.28 × 10-4 wb/A-m. Find its relative permeability and susceptibility ?

Sol: $\large \mu_r = \frac{\mu}{\mu_0}$

$\large \mu_r = \frac{6.28 \times 10^{-4}}{4\pi \times 10^{-7}}$

μr = 500

$\large \mu_r = 1 + \chi$

$\large \chi = \mu_r – 1 = 500 -1$

χ = 499.

A magnetising field of 1600 Am^(-1) produces a magnetic flux of 2.4 × 10^(-5) weber in a bar of iron…

Q: A magnetising field of 1600 Am-1 produces a magnetic flux of 2.4 × 10-5 weber in a bar of iron of cross section 0.2 cm2. Calculate permeability and susceptibility of the bar.

Sol: Magnetic induction $\large B = \frac{\phi}{A} = \frac{2.4 \times 10^{-5}}{0.2 \times 10^{-4}}$

B = = 1.2 Wb/m2

(i) Permeability, μ = B/H = 1.2/1600 = 7.5 × 10-4  TA-1m

(ii) As μ = μ0 (1 + χ) then

Susceptibility, $\large \chi = \frac{\mu}{\mu_0} – 1$

$\large \chi = \frac{7.5 \times 10^{-4}}{4\pi \times 10^{-7}} – 1$

= 596.1

Two bar magnets of the same length and breadth but having magnetic moments M and 2M are joined with like poles together…

Q: Two bar magnets of the same length and breadth but having magnetic moments M and 2M are joined with like poles together and suspended by a string. The time of oscillation of this assembly in a magnetic field of strength B is 3 sec. What will be the period of oscillation, if the polarity of one of the magnets is changed and the combination is again made to oscillate in the same field ?

Sol: As magnetic moment is a vector, so when magnets are joined with like poles together

M1 = M + 2 M

M1 = 3 M

$\large T = 2 \pi \sqrt{\frac{I_1 + I_2}{3 M B}}$ ….(i)

When the polarity of the one of the magnets is reversed, M2 = M ~ 2 M = M

$\large T’ = 2 \pi \sqrt{\frac{I_1 + I_2}{M B}}$ …..(ii)

Dividing eq. (ii) by (i),

$\large \frac{T’}{T} = \sqrt{3}$

$\large T’ = \sqrt{3} T = 3 \sqrt{3} sec$

When a short bar magnet is kept in tan A position on a deflection magnetometer, the magnetic needle oscillates with…

Q: When a short bar magnet is kept in tan A position on a deflection magnetometer, the magnetic needle oscillates with a frequency ‘ f ’ and the deflection produced is 45°. If the bar magnet is removed find the frequency of oscillation of that needle?

Sol: $\large \nu \propto \sqrt{B}$

$\large \frac{\nu_1}{\nu_2} = \sqrt{\frac{B_1}{B_2}}$

Where $\large B_1 = \sqrt{B^2 + B_H^2}$

$\large B_1 = \sqrt{(B_H tan45^o)^2 + B_H^2} = \sqrt{2}B_H$

And , B2 = BH

$\large \frac{\nu_1}{\nu_2} = \sqrt{\frac{\sqrt{2}B_H}{B_H}} = 2^{1/4}$

$\large n_2 = \frac{n_1}{2^{1/4}} = \frac{f}{2^{1/4}}$