A compass needle of magnetic moment $60 {Am}^{2}$ , pointing towards geographical north at a certain place where the horizontal component of earth's magnetic field is $40 μ {wb m}^{2}$ experiences a torque of $1.2 \times 10^{- 3} N-m$ . The declination at that place _____ (in degree) (Give integer value).

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Solution

Torque , $T = {MB}_{H} sin θ$ where $θ$ is the declination angle.

$∴ sin θ = \frac{T}{M B_{H}}$

Substituting the given data we get,

$sin θ = \frac{1.2 \times 10^{- 3}}{60 \times 40 \times 10^{- 6}} = \frac{12}{24} = \frac{1}{2}$

$\Rightarrow θ$ = $30^{o}$ .

Accepted answer : $30$

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Similar questions

Q. A compass needle whose magnetic moment is

60

m^{2}

pointing geographical north at a certain place where the horizontal component of earth's magnetic field is

40 \times 10^{- 6}

m^{- 2}

experiences a torque of

1.2 \times 10^{- 3}

N m. The declination of the place is?

Q. At a certain location in Africa, compass points

12^{0}

west of geographic north. The north tip of magnetic needle of a dip circle placed in the plane of magnetic meridian points

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above the horizontal. The horizontal component of earth's field is measured to be

0.16

G. The magnitude of earth's field at the location is :

At a certain location in Africa, a compass points 12º west of the geographic north. The north tip of the magnetic needle of a dip circle placed in the plane of magnetic meridian points 60º above the horizontal. The horizontal component of the earth’s field is measured to be 0.16 G. Specify the direction and magnitude of the earth’s field at the location.

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A compass needle of magnetic moment 60 Am2, pointing towards geographical north at a certain place where the horizontal component of earth's magnetic field is 40μwb m2 experiences a torque of 1.2×10−3 N-m . The declination at that place _____ (in degree) (Give integer value).

Torque ,T=MBHsinθwhere θ is the declination angle. ∴sinθ=TMBH Substituting the given data we get, sinθ=1.2×10−360×40×10−6=1224=12 ⇒θ = 30o. Accepted answer : 30

Torque , $T = {MB}_{H} sin θ$ where $θ$ is the declination angle.

$∴ sin θ = \frac{T}{M B_{H}}$

Substituting the given data we get,

$sin θ = \frac{1.2 \times 10^{- 3}}{60 \times 40 \times 10^{- 6}} = \frac{12}{24} = \frac{1}{2}$

$\Rightarrow θ$ = $30^{o}$ .

Accepted answer : $30$