A compass needle whose magnetic moment is 60 A m 2 pointing geographical north at a certain place, where the horizontal component of earth's magnetic field is 40 μ Wb / m 2, experiences a torque 1.2 × 10-3 N m. What is the declination at this place?

Question

A compass needle whose magnetic moment is 60 A m 2 pointing geographical north at a certain place, where the horizontal component of earth's magnetic field is 40 μ Wb / m 2, experiences a torque 1.2 × 10-3 N m. What is the declination at this place? (A) 30° (B) 45° (C) 60° (D) 25°

Tardigrade Admin · Accepted Answer

As the compass needle is free to rotate in a horizontal plane and points along the magnetic meridian,

so when it is pointing along the geographic meridian, it will experience a torque due to the horizontal component of earth's magnetic field,
i.e.,

τ = M B_{H} sin θ

where

θ =

angle between geographical and magnetic meridians called angle of declination
So,

sin θ = \frac{1.2 \times 1 0 ^{- 3}}{60 \times 40 \times 1 0 ^{- 6}}

= \frac{1}{2} \Rightarrow θ = 3 0^{\circ}

Q. A compass needle whose magnetic moment is $60 A m^{2}$ pointing geographical north at a certain place, where the horizontal component of earth's magnetic field is $40 μ Wb / m^{2}$ , experiences a torque $1.2 \times 1 0^{- 3} N m$ . What is the declination at this place?

$3 0^{\circ}$

$4 5^{\circ}$

$6 0^{\circ}$

$2 5^{\circ}$

Q. A compass needle whose magnetic moment is 60Am2 pointing geographical north at a certain place, where the horizontal component of earth's magnetic field is 40μWb/m2, experiences a torque 1.2×10−3Nm. What is the declination at this place?

30∘

45∘

60∘

25∘

Q. A compass needle whose magnetic moment is $60 A m^{2}$ pointing geographical north at a certain place, where the horizontal component of earth's magnetic field is $40 μ Wb / m^{2}$ , experiences a torque $1.2 \times 1 0^{- 3} N m$ . What is the declination at this place?

$3 0^{\circ}$

$4 5^{\circ}$

$6 0^{\circ}$

$2 5^{\circ}$