Question

The magnetic moment of the assumed dipole at the earth's centre is 8.0 × 1022 A m2. Calculate the magnetic field B at the geomagnetic poles of the earth. Radius of the earth is 6400 km.

Answer 1

Explanation:

To find: The minimum value of v so that the vertically-upward component of velocity is non-positive for each photoelectron.

Step 1:

Given data in the question  

Magnetic dipolar moment at the center of the Earth,  $M=8.0 \times 10^{22} \mathrm{Am}^{2}$

Earth Radius, d = 6400 km

At the end of the Earth's position (axial) is the geomagnetic pole.

The magnetic field at the dipole's axial level

Step 2:

We know that  

$B=\frac{\mu_{0} M}{4 \pi d^{3}}$

$B=\frac{10^{-7} \times 2 \times 8 \times 10^{22}}{64^{3} \times 10^{15}}$

$B=\frac{10^{-7} \times 16 \times 10^{22}}{64^{3} \times 10^{15}}$

$B=\frac{10^{-7} \times 16 \times 10^{22}}{262144 \times 10^{15}}$

$B=\frac{16 \times 10^{15}}{262144 \times 10^{15}}$

$B=\frac{16}{262144}$

$\begin{aligned}&B=6.1 \times 10^{-5} T\\&B=61 \times 10^{-6} T\end{aligned}$

B = 61 μT