Ask & Answer: India's Largest Education Community

5.13 Earth’s magnetic field at the given place, H = 0.36 G

The magnetic field at a distance d, on the axis of the magnet is given as

$B_1$ = $\frac{{\mu }_0}{4\pi }\frac{2M}{d^3}$ = H ……………….(1)

where ${\mu }_0$ = Permeability of free space = 4 $\pi \times {10}^{-7}$ T m $A^{-1}$ and M = magnetic moment

Total magnetic field, B = $B_1+B_2$ = H + $\frac{H}{2}$ = 0.36 + 0.18 = 0.54 G

Therefore, the magnetic field is 0.54 G in the direction of earth’s magnetic field.

The CUET PG cutoffs are released individually by each participating university or college. The CUET PG Cutoff represent the minimum marks and percentiles required for admission in the specific postgraduate courses in each institute.

The CUET PG Cutoff for top programs in the central universities, for the general category candidates, is expected to be 220-250 marks, and for the reserved category candidates, it is around 180-210 marks, as per the course type and institution.

5.12 Magnetic moment of the bar magnet, M = 0.48 J/T

The distance, d = 10 cm = 0.1 m

The magnetic field at a distance d from the centre of the magnet on the axis is given by the relation:

B = $\frac{{\mu }_0}{4\pi }\frac{2M}{d^3}$ , where ${\mu }_0$ = Permeability of free space = 4 $\pi \times {10}^{-7}$ T m $A^{-1}$ $\pi \times {10}^{-7}$

B = $\frac{4\pi \times {10}^{-7}\times 2\times 0.48}{4\pi \times {0.1}^3}$ = 0.96 $\times {10}^{-6}$ T = 0.96 G

The magnetic field is along S – N direction.

The distance, d = 10 cm = 0.1 m

The magnetic field at a distance d from the centre of the magnet on the equatorial line of the magnet is given by the relation:

B = $\frac{{\mu }_0}{4\pi }\frac{M}{d^3}$ , where ${\mu }_0$ = Permeability of free space = 4 $\pi \times {10}^{-7}$ T m $A^{-1}$

B = $\frac{4\pi \times {10}^{-7}\times 0.48}{4\pi \times {0.1}^3}$ = 0.48 T = 0.48 $\times {10}^{-6}$ G

The magnetic field is along N – S direction.

5.11 Given:

The angle of declination, $\theta$ = 12 $°$

The angle of dip, $\beta$ = 60 $°$

Horizontal component of Earth’s magnetic field, $B_H$ = 0.16 G

If Earth’s magnetic field be B, we can relate B and $B_H$ as $B_H=B\cos ?\beta$

B = $\frac{B_H}{\cos ?\beta }$ = $\frac{0.16}{\cos ?60°}$ = 0.32 G

Therefore, the Earth’s magnetic field lies in the vertical plane, 12west of the geographic meridian, making an angle of 60 $°$ (upward) with the horizontal direction. Its magnitude is 0.32 G.

5.10 Given:

Horizontal component of Earth’s magnetic field, $B_H$ = 0.35 G

Angle made by the needle with the horizontal plane, $\beta$ = 22 $°$

If the Earth’s magnetic field at that location be B,

then $B_H$ = B $\cos ?\beta$

B = $\frac{B_H}{\cos ?\beta }$ = $\frac{0.35}{\cos ?22°}$ = 0.377 G

Therefore, the strength of Earth’s magnetic field at that location is 0.377 G

5.9 Number of turns, n = 16

Radius of the coil, r = 10 cm = 0.1 m

Cross-section of the coil, A = $\pi r^2$ = $\pi \times {0.1}^2$ $m^2$ = 0.0314 $m^2$

Current in the coil, I = 0.75 A

Magnetic field strength, B = 5.0 $\times {10}^{-2}$ T

Frequency of oscillation of the coil, $\nu$ = 2/s

Magnetic moment, M = NIA = 16 $\times 0.75\times 0.0314$ = 0.377 A $m^2$

Frequency is given by the relation, $\nu$ = $\frac{1}{2\pi }\sqrt{\frac{MB}{I}}$ , where I = moment of Inertia of the coil

I = $\frac{MB}{4{\pi }^2{\nu }^2}$ = $\frac{0.377\mathrm{}\times 5.0\mathrm{}\times {10}^{-2}}{4{\pi }^2\times 2^2}$ = 1.19 $\times {10}^{-4}$ kg $m^2$

5.8 (a) Number of turns, n = 2000

Area of cross-section, A = 1.6 $\times {10}^{-4}$ $m^2$

Current, I = 4.0 A

The magnetic moment along the axis of the solenoid is calculated as

M = nAI = 2000 $\times$ 1.6 $\times {10}^{-4}\times 4$ = 1.28 A $m^2$

(b) Magnetic field, B = 7.5 $\times {10}^{-2}$ T

Angle between magnetic field and the axis of the solenoid, $\theta$ = 30 $°$

Torque $\tau =MB\sin ?\theta$

= 1.28 $\times$ 7.5 $\times {10}^{-2}$ $\times \sin ?30°$

= 4.8 $\times {10}^{-2}$ Nm

5.6 Magnetic field strength, B = 0.25 T

Magnetic moment, M = 0.6 J/T

The angle $\theta$ , between the axis of the solenoid and the direction of the applied field is 30 $°$ . Therefore, the torque acting on the solenoid is given as

$\tau =MBsin\theta$ = 0.6 $\times 0.25\times \sin ?30°$ = 7.5 $\times {10}^{-2}$ J

The chances are very low to get the ECE branch in Sastra, as the cutoff for the ECE branch this year was about 988 in IPE alone (stream 2). But in stream 1 (ipe+ mains), you should have about 95 percentile in mains for those ipe marks.

Hi, yes, since the hospitality industry uses English as a working language, proficiency in English is essential.

Hi, programs like B.Sc. in Hospitality and Hotel Administration, Bachelor of Hotel Management (BHM), and various diploma/PG Diploma courses are offered by top Hotel Management colleges in West Bengal. 

The TNEA 2025 Cutoff will be released after each round of counselling, which is expected to start on July 14th and will continue till August 19th, 2025. The DoTE also released the TNEA 2025 Rank list on 27th June 2025.

The CSE branch cutoff is expected to be between 198 and 200 marks for candidates in the general category, between 195 and 199 for ECE, and between 192-197 for mechanical for major universities like Anna University.

Candidates from BC and MBC can get into the CSE Branch at top colleges such as CEG and PSG with scores ranging from 192 to 195, while candidates from SC can obtain the same with scores between 185 and 190.