Earth's Magnetism — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Earth's Magnetism MCQs with step-by-step solutions (20 questions). Part of Magnetic Materials. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Earth's Magnetism · easy · theory
The three elements of the Earth's magnetic field are:
A. Declination, latitude and longitude
B. Dip, latitude and the vertical component
C. Susceptibility, permeability and flux
D. Declination, dip and the horizontal component ✓ Correct
Solution: Together these three quantities fix the magnitude and direction of the field at any place.
Q2 — Earth's Magnetism · medium · theory
Magnetic declination at a place is the angle between:
A. The geographic meridian and the magnetic meridian ✓ Correct
B. The Earth's field and the horizontal
C. The magnetic axis and the rotation axis of a compass needle
D. The horizontal and vertical components
Solution: It tells how far a compass needle points away from true geographic north.
Q3 — Earth's Magnetism · medium · theory
The angle of dip at a place is the angle between:
A. The Earth's total field and the horizontal direction ✓ Correct
B. The vertical and the magnetic meridian
C. The geographic and magnetic meridians
D. Two successive magnetic meridians
Solution: A dip needle free to rotate in the vertical magnetic meridian settles at this angle.
Q4 — Earth's Magnetism · easy · theory
The angle of dip at the magnetic equator and at the magnetic poles is respectively:
A. $0^\circ$ and $90^\circ$ ✓ Correct
B. $0^\circ$ and $0^\circ$
C. $90^\circ$ and $0^\circ$
D. $45^\circ$ and $90^\circ$
Solution: At the equator the Earth's field is horizontal; at the poles it is vertical.
Q5 — Earth's Magnetism · medium · theory
The horizontal and vertical components of the Earth's field of magnitude $B$ at dip angle $\delta$ are:
A. $B$ and $B\tan\delta$
B. $B\cos\delta$ and $B\sin\delta$ ✓ Correct
C. $B\sin\delta$ and $B\cos\delta$
D. $B\tan\delta$ and $B$
Solution: The dip angle is measured from the horizontal, so the horizontal component carries the cosine.
Q6 — Earth's Magnetism · medium · theory
The angle of dip $\delta$ is related to the components of the Earth's field by:
A. $\tan\delta = \dfrac{B_H}{B_V}$
B. $\sin\delta = \dfrac{B_V}{B_H}$
C. $\cos\delta = \dfrac{B_V}{B_H}$
D. $\tan\delta = \dfrac{B_V}{B_H}$ ✓ Correct
Solution: Dividing $B_V = B\sin\delta$ by $B_H = B\cos\delta$ gives the tangent.
Q7 — Earth's Magnetism · medium · theory
The magnetic equator is the locus of places where the:
A. Total field is zero
B. Horizontal component is zero
C. Angle of declination is zero
D. Angle of dip is zero ✓ Correct
Solution: It lies close to, but does not coincide with, the geographic equator.
Q8 — Earth's Magnetism · hard · theory
A neutral point near a bar magnet is a point where:
A. The dip angle is $90^\circ$
B. The field of the magnet exactly cancels the Earth's horizontal field ✓ Correct
C. The Earth's field alone is zero
D. The magnet exerts no force on any material
Solution: A compass placed there shows no definite direction, since the resultant horizontal field is zero.
Q9 — Earth's Magnetism · easy · numerical
At a place the angle of dip is $0^\circ$. The vertical component of the Earth's field there is:
A. Equal to the horizontal component
B. Equal to the total field
C. Zero ✓ Correct
D. Maximum
Solution: $B_V = B\sin 0^\circ = 0$; the field is entirely horizontal, as at the magnetic equator.
Q10 — Earth's Magnetism · easy · numerical
At a place the angle of dip is $90^\circ$. The horizontal component of the Earth's field there is:
A. Half the total field
B. Zero ✓ Correct
C. Maximum
D. Equal to the total field
Solution: $B_H = B\cos 90^\circ = 0$; the field is entirely vertical, as at a magnetic pole.
Q11 — Earth's Magnetism · medium · numerical
The Earth's total field at a place is $5 \times 10^{-5}\text{ T}$ and the dip is $60^\circ$. The horizontal component is:
A. $2.5 \times 10^{-5}\text{ T}$ ✓ Correct
B. $4.33 \times 10^{-5}\text{ T}$
C. $8.66 \times 10^{-5}\text{ T}$
D. $5 \times 10^{-5}\text{ T}$
Solution: $B_H = B\cos 60^\circ = 5 \times 10^{-5} \times 0.5 = 2.5 \times 10^{-5}\text{ T}$.
Q12 — Earth's Magnetism · medium · numerical
For the same place ($B = 5 \times 10^{-5}\text{ T}$, dip $60^\circ$), the vertical component is:
A. $5 \times 10^{-5}\text{ T}$
B. $4.33 \times 10^{-5}\text{ T}$ ✓ Correct
C. $1.25 \times 10^{-5}\text{ T}$
D. $2.5 \times 10^{-5}\text{ T}$
Solution: $B_V = B\sin 60^\circ = 5 \times 10^{-5} \times 0.866 \approx 4.33 \times 10^{-5}\text{ T}$.
Q13 — Earth's Magnetism · medium · numerical
At a place the vertical and horizontal components of the Earth's field are equal. The angle of dip is:
A. $30^\circ$
B. $60^\circ$
C. $45^\circ$ ✓ Correct
D. $90^\circ$
Solution: $\tan\delta = \dfrac{B_V}{B_H} = 1$, so $\delta = 45^\circ$.
Q14 — Earth's Magnetism · hard · numerical
At a place $B_H = 3 \times 10^{-5}\text{ T}$ and $B_V = 4 \times 10^{-5}\text{ T}$. The total field is:
A. $3.5 \times 10^{-5}\text{ T}$
B. $1 \times 10^{-5}\text{ T}$
C. $7 \times 10^{-5}\text{ T}$
D. $5 \times 10^{-5}\text{ T}$ ✓ Correct
Solution: $B = \sqrt{B_H^2 + B_V^2} = \sqrt{(3)^2 + (4)^2} \times 10^{-5} = 5 \times 10^{-5}\text{ T}$.
Q15 — Earth's Magnetism · hard · numerical
At a place $B_V = 4 \times 10^{-5}\text{ T}$ and $B_H = 3 \times 10^{-5}\text{ T}$. The angle of dip is approximately:
A. $45^\circ$
B. $36.9^\circ$
C. $53.1^\circ$ ✓ Correct
D. $60^\circ$
Solution: $\tan\delta = \dfrac{4}{3} = 1.333$, so $\delta \approx 53.1^\circ$.
Q16 — Earth's Magnetism · hard · numerical
The Earth's field at a place is $6 \times 10^{-5}\text{ T}$ with a dip of $30^\circ$. The horizontal component is approximately:
A. $5.2 \times 10^{-5}\text{ T}$ ✓ Correct
B. $3 \times 10^{-5}\text{ T}$
C. $6 \times 10^{-5}\text{ T}$
D. $1.5 \times 10^{-5}\text{ T}$
Solution: $B_H = 6 \times 10^{-5} \times \cos 30^\circ = 6 \times 10^{-5} \times 0.866 \approx 5.2 \times 10^{-5}\text{ T}$.
Q17 — Earth's Magnetism · medium · numerical
At the magnetic equator, the Earth's total field equals:
A. Twice its horizontal component
B. Zero
C. Its vertical component
D. Its horizontal component ✓ Correct
Solution: With $\delta = 0$, $B_V = 0$ and $B = B_H$.
Q18 — Earth's Magnetism · medium · numerical
At a magnetic pole, the Earth's total field equals:
A. Its vertical component ✓ Correct
B. Half its vertical component
C. Zero
D. Its horizontal component
Solution: With $\delta = 90^\circ$, $B_H = 0$ and $B = B_V$.
Q19 — Earth's Magnetism · hard · numerical
At a place $B_H = 3.4 \times 10^{-5}\text{ T}$ and the total field is $6.8 \times 10^{-5}\text{ T}$. The angle of dip is:
A. $90^\circ$
B. $45^\circ$
C. $60^\circ$ ✓ Correct
D. $30^\circ$
Solution: $\cos\delta = \dfrac{B_H}{B} = \dfrac{3.4}{6.8} = 0.5$, so $\delta = 60^\circ$.
Q20 — Earth's Magnetism · medium · numerical
At a place the ratio $\dfrac{B_V}{B_H}$ is $\dfrac{1}{\sqrt{3}}$. The angle of dip is:
A. $45^\circ$
B. $60^\circ$
C. $15^\circ$
D. $30^\circ$ ✓ Correct
Solution: $\tan\delta = \dfrac{1}{\sqrt{3}}$, so $\delta = 30^\circ$.