Question

The magnetometer of the previous problem is used with the same magnet in Tan-B position. Where should the magnet be placed to produce a 37° deflection of the needle?

Answer 1

The magnet is 7.9 cm away from the centre.

Explanation:

Step 1:

Given data in the question  

Deflection in the magnetometer at the given position when put in a short magnet's magnetic field,

θ = 37°

distance between the needle and the magnet, d = 10 cm = 0.1 m

Let M be the magnetic moment of that magnet, and let $B_H$ be the magnetic field parallel to Earth.

Step 2:

The magnetometer theory says,

$\frac{M}{B_{H}}=\frac{4 \pi}{\mu_{0}} \frac{\left(d^{2}-l^{2}\right)^{2}}{2 d} \tan \theta$

For the Short Magnet ,

$\frac{M}{B_{H}}=\frac{4 \pi}{\mu_{0}} \frac{d^{4}}{2 d} \tan \theta$

$\frac{M}{B_{H}}=\frac{4 \pi}{4 \pi \times 10^{-7}} \times \frac{0.1^{3}}{2} \times \tan 37^{\circ}$

$\frac{M}{B_{H}}=0.5 \times 0.75 \times 1 \times 10^{4}$

$\frac{M}{B_{H}}=3.75 \times 10^{3} A-m^{2} / T$

The magnetometer is deflective θ = 37°

Step 3:

From the theory of magnetometers in place Tan-B we have

$\frac{M}{B_{H}}=\frac{4 \pi}{\mu_{0}}\left(d^{2}+l^{2}\right)^{\frac{3}{2}} \tan \theta$

As we can ignore l w.r.t. d for the small magnet l<<<d.

$\frac{M}{B_{H}}=\frac{4 \pi}{\mu_{0}} d^{3} \tan \theta$

$3.75 \times 10^{3}=\frac{1}{10^{-7}} \times d^{3} \times 0.75$

$d^{3}=\frac{3.75 \times 10^{3} \times 10^{-7}}{0.75}=5 \times 10^{-4}$

$d=\sqrt[3]{5 \times 10^{-4}}$

$d=0.079 \mathrm{m}=7.9 \mathrm{cm}$

The magnet is 7.9 cm away from the centre.