When placed at a separation of 10 cm from the needle, a short magnet causes a deflection of 37^ ^ ^ in a deflection magnetometer in the Tan-A position. Determine the ratio of the magnet's magnetic moment to the earth's

Question

Quantrex Academy · Accepted Answer

Correct answer: D. 3750 A ^2/T

In the Tan-A position, the magnetic field

B

due to a short magnet on its axial line is given by:

B = \dfrac{\mu_0}{4\pi} \dfrac{2M}{d^3}

The deflection

\theta

is given by the relation:

B = B_H \tan\theta

Equating the two expressions for

B

:

\dfrac{\mu_0}{4\pi} \dfrac{2M}{d^3} = B_H \tan\theta

Rearranging to find the ratio

\dfrac{M}{B_H}

:

\dfrac{M}{B_H} = \dfrac{d^3 \tan\theta}{2 \left( \dfrac{\mu_0}{4\pi} \right)}

Substituting

d = 10 \text{ cm} = 0.1 \text{ m}

,

\theta = 37^{}^{{}^{\circ}}

, and

\dfrac{\mu_0}{4\pi} = 10^{-7} \text{ T}\cdot\text{m/A}

:

\dfrac{M}{B_H} = \dfrac{(0.1)^3 \times \tan 37^{}^{{}^{\circ}}}{2 \times 10^{-7}}

Using

\tan 37^{}^{{}^{\circ}} = \dfrac{3}{4} = 0.75

:

\dfrac{M}{B_H} = \dfrac{10^{-3} \times 0.75}{2 \times 10^{-7}}

\dfrac{M}{B_H} = \dfrac{0.75}{2} \times 10^4

\dfrac{M}{B_H} = 0.375 \times 10000 = 3750 \text{ A}\cdot\text{m}^2/\text{T}

When placed at a separation of 10 cm from the needle, a short...

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