JEE Physics Magnetism and Matter MCQs Set 05

Answer: (d) \( \cot^2\theta=\cot^2\theta_1+\cot^2\theta_2 \)

Explanation:
1. Let the true dip be \( \theta \) with horizontal component \( H \) and vertical component \( V \). In a vertical plane making angle \( \alpha \) with the meridian, \( \cot\theta'=\frac{H\cos\alpha}{V}=\cot\theta\cos\alpha \).
2. For two perpendicular planes, \( \cot\theta_1=\cot\theta\cos\alpha \) and \( \cot\theta_2=\cot\theta\sin\alpha \).
3. Squaring and adding: \( \cot^2\theta=\cot^2\theta_1+\cot^2\theta_2 \).

Teacher's Note:
1. Apparent dip needle readings in vertical planes measure only a component of H. Adding the squares of the two perpendicular components recovers the true value.

Answer: (d) low retentivity and low coercivity

Explanation:
1. An electromagnet must be magnetised easily and lose its magnetism when the current is switched off.
2. Low retentivity means little magnetism is retained, and low coercivity means it is easily demagnetised.
3. So soft iron, with low retentivity and low coercivity, is used.

Teacher's Note:
1. Soft iron has a narrow hysteresis loop, so very little energy is wasted in reversing the field, which is ideal for electromagnets and transformer cores.

Answer: (a) Ni

Explanation:
1. Gold, carbon (graphite) and copper are diamagnetic.
2. Nickel is a ferromagnetic material.
3. So Ni is the one that is not diamagnetic.

Teacher's Note:
1. Common diamagnetic materials are Cu, Au, Ag, Bi, water and graphite. Fe, Co and Ni are ferromagnetic.

Answer: (b) 2

Explanation:
1. Flux density inside the material \( B=\mu H \) and outside (in the surroundings) \( B_0=\mu_0H \).
2. \( \frac{B}{B_0}=\frac{\mu}{\mu_0}=\mu_r=2 \).
3. The relative permeability is 2.

Teacher's Note:
1. The ratio of the field inside the material to the applied field is the relative permeability.

Answer: (b) ferromagnetic

Explanation:
1. For diamagnetic and paramagnetic materials the susceptibility is nearly independent of the applied field strength (paramagnets depend on temperature only).
2. Only a ferromagnetic material has a susceptibility that varies with both temperature and the applied field.
3. So the material is ferromagnetic.

Teacher's Note:
1. Ferromagnets show non-linear magnetisation and hysteresis, so their susceptibility is not a constant.

Answer: (b) a broad hysteresis loop

Explanation:
1. A hard ferromagnetic material retains its magnetisation strongly (high retentivity) and needs a large reverse field to demagnetise it (high coercivity).
2. This means a wide B-H loop enclosing a large area.
3. So it has a broad hysteresis loop.

Teacher's Note:
1. The area of the hysteresis loop equals the energy lost per cycle. Hard magnetic materials have large loops.

Answer: (d) cobalt-steel

Explanation:
1. Permanent magnets need high retentivity and high coercivity.
2. Alloys such as cobalt-steel, alnico and ticonal have these properties.
3. So powerful permanent magnets are made of cobalt-steel.

Teacher's Note:
1. Alloys of iron with cobalt, aluminium and nickel make the strongest permanent magnets.

Answer: (d) high retentivity and high coercivity

Explanation:
1. A permanent magnet must retain its magnetism for a long time (high retentivity) and not be demagnetised by stray fields (high coercivity).
2. So the material has high retentivity and high coercivity.

Teacher's Note:
1. Permanent magnets need high retentivity and high coercivity. Electromagnets need low values of both.

Answer: (a) 0.33 J

Explanation:
1. Potential energy \( U=-MB\cos\theta \). Aligned: \( U=-MB \). Normal: \( U=0 \).
2. \( W=\Delta U=0-(-MB)=MB \).
3. \( W=1.5\times0.22=0.33 \) J.

Teacher's Note:
1. Work done on a dipole equals the increase in its potential energy. Turning through 90 degrees costs MB, and through 180 degrees costs 2MB.

Answer: (a) \( 1.05\times10^{23} \) A m²

Explanation:
1. At the equatorial position \( B_E=\frac{\mu_0}{4\pi}\frac{M}{r^3} \).
2. \( M=\frac{B_E r^3}{\mu_0/4\pi}=\frac{4\times10^{-5}\times(6.4\times10^6)^3}{10^{-7}} \).
3. \( M\approx1.05\times10^{23} \) A m².

Teacher's Note:
1. Convert gauss to tesla first (1 G = 10^-4 T). The factor mu0 over 4 pi equals 10^-7 in SI units.

Answer: (a) 198

Explanation:
1. \( B=\frac{\phi}{A}=\frac{25\times10^{-6}}{0.25\times10^{-4}}=1 \) Wb/m². Permeability \( \mu=\frac BH=\frac1{4000}=2.5\times10^{-4} \).
2. \( \mu_r=\frac{\mu}{\mu_0}=\frac{2.5\times10^{-4}}{4\pi\times10^{-7}}\approx199 \).
3. \( \chi=\mu_r-1\approx198 \).

Teacher's Note:
1. Get B from flux and area, then permeability from B over H, then relative permeability. Susceptibility is one less than relative permeability.

Answer: (a) \( 90^\circ \) and \( 0^\circ \)

Explanation:
1. The angle of dip is the angle the earth's total field makes with the horizontal.
2. A needle pointing vertically means the field is vertical: dip \( =90^\circ \) (a magnetic pole).
3. A needle pointing horizontally means dip \( =0^\circ \) (the magnetic equator).

Teacher's Note:
1. Dip is 0 degrees at the magnetic equator and 90 degrees at the magnetic poles.