Earth's magnetic field and magnetic elements — NEET & JEE Questions

Question 1 · difficulty L1 · recall

The angle between geographic north and magnetic north in the horizontal plane is called

• A. angle of dip
• B. magnetic inclination only
• C. magnetic declination
• D. latitude

Question 2 · difficulty L1 · understanding

The angle made by Earth's magnetic field with the horizontal is called

• A. declination
• B. azimuth only
• C. colatitude
• D. angle of dip or inclination

Question 3 · difficulty L2 · understanding

If Earth's field magnitude is B and dip is I, the horizontal component is

• A. B cos I
• B. B sin I
• C. B tan I
• D. B/cos I

Question 4 · difficulty L2 · application

If Earth's field magnitude is B and dip is I, the vertical component magnitude is

• A. B cos I
• B. B sin I
• C. B cot I
• D. B/sin I

Question 5 · difficulty L2 · understanding

The vertical component of the earth's magnetic field is $$6 \times 10^{-5} \mathrm{~T}$$ at any place where the angle of dip is $$37^{\circ}$$. The earth's resultant magnetic field at that place will be $$\left(\right.$$Given $$\left.\tan 37^{\circ}=\frac{3}{4}\right)$$

• A. $$8 \times 10^{-5} \mathrm{~T}$$
• B. $$6 \times 10^{-5} \mathrm{~T}$$
• C. $$5 \times 10^{-4} \mathrm{~T}$$
• D. $$1 \times 10^{-4} \mathrm{~T}$$

Question 6 · difficulty L2 · understanding

An electron with energy 0.1 keV moves at right angle to the earth's magnetic field of 1 $$\times$$ 10 $$-$$4 Wbm $$-$$2 . The frequency of revolution of the electron will be : (Take mass of electron = 9.0 $$\times$$ 10 $$-$$31 kg)

• A. $$1.6 \times 10^{5} \mathrm{~Hz}$$
• B. $$5.6 \times 10^{5} \mathrm{~Hz}$$
• C. $$2.8 \times 10^{6} \mathrm{~Hz}$$
• D. $$1.8 \times 10^{6} \mathrm{~Hz}$$

Question 7 · difficulty L2 · understanding

At a certain place the angle of dip is 30$$^\circ$$ and the horizontal component of earth's magnetic field is 0.5 G. The earth's total magnetic field (in G), at that certain place, is :

• A. $${1 \over {\sqrt 3 }}$$
• B. $${1 \over 2}$$
• C. $$\sqrt 3$$
• D. 1

Question 8 · difficulty L3 · application

At a place where the dip is 60° and total field is B, the horizontal component equals

• A. sqrt(3)B/2
• B. 2B
• C. B/2
• D. B/sqrt(2)

Question 9 · difficulty L3 · application

If BH and BV are the horizontal and vertical field components, tan I equals

• A. BH/BV
• B. B/BH
• C. BH·BV
• D. BV/BH

Question 10 · difficulty L3 · understanding

A compass needle of oscillation magnetometer oscillates 20 times per minute at a place $$\mathrm{P}$$ of $$\operatorname{dip} 30^{\circ}$$. The number of oscillations per minute become 10 at another place $$\mathrm{Q}$$ of $$60^{\circ}$$ dip. The ratio of the total magnetic field at the two places $$\left(B_{Q}: B_{P}\right)$$ is :

• A. $$\sqrt{3}: 4$$
• B. $$4: \sqrt{3}$$
• C. $$\sqrt{3}: 2$$
• D. $$2: \sqrt{3}$$