Chemistry, asked by Nadeemsheikh8474, 10 months ago

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.

Answers

Answered by bhuvna789456
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

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