The earth's magnetic field at the geometric poles is √10 × 10-5 T. The magnitude of the field at a point on the earth's surface where the radius makes an angle θ with the axis of earth's assumed magnetic dipole is 5 × 10-5 T. The magnitude of θ in degree is :

Question

The earth's magnetic field at the geometric poles is √10 × 10-5 T. The magnitude of the field at a point on the earth's surface where the radius makes an angle θ with the axis of earth's assumed magnetic dipole is 5 × 10-5 T. The magnitude of θ in degree is : (A) 30° (B) 60° (C) 45° (D) 75°

Tardigrade Admin · Accepted Answer

According to the question, Bv=√10 × 10-5 T <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/d1899d49c79d72bfe91be3db0f1c688b-.png" /> In Δ OBC, ∵ cos θ=(BV/RR) =(√10 × 10-5/5 × 10-5) =(√2/√5) or θ=50.76° ≈ 45° Hence, the magnitude of θ is 45°.

Q. The earth's magnetic field at the geometric poles is $10 \times 1 0^{- 5} T$ . The magnitude of the field at a point on the earth's surface where the radius makes an angle $θ$ with the axis of earth's assumed magnetic dipole is $5 \times 1 0^{- 5} T$ . The magnitude of $θ$ in degree is :

$3 0^{\circ}$

$6 0^{\circ}$

$4 5^{\circ}$

$7 5^{\circ}$

According to the question,
$B_{v} = 10 \times 1 0^{- 5} T$

In $Δ OBC$ ,
$∵ cos θ = \frac{B _{V}}{R _{R}}$
$= \frac{10 \times 1 0 ^{- 5}}{5 \times 1 0 ^{- 5}}$
$= \frac{2}{5}$
or $θ = 50.7 6^{\circ} \approx 4 5^{\circ}$
Hence, the magnitude of $θ$ is $4 5^{\circ}$ .

Q. The earth's magnetic field at the geometric poles is 10​×10−5T. The magnitude of the field at a point on the earth's surface where the radius makes an angle θ with the axis of earth's assumed magnetic dipole is 5×10−5T. The magnitude of θ in degree is :

30∘

60∘

45∘

75∘