Question

A proton is circling the Earth above the magnetic equator, where Earth’s magnetic field is directed horizontally north and has a magnitude of 4.00 × 10–8 T. If the proton is moving at a speed of 2.7 × 107 m/s, how far above the surface of the Earth is the proton

Answer 1

Answer:

$6.65\times 10^5 m$

Explanation:

We are given that

Magnetic field=B=$4\times 10^{-8} T$

$v=2.7\times 10^7 m/s$

We have to find the height of proton from the surface of the Earth.

Mass of proton,$m_p=1.67\times 10^{-27} kg$

Charge on proton,$q=1.6\times 10^{-19} C$

Radius of Earth, r=$6.38\times 10^6 m$

Centripetal force due to rotation of proton=$\frac{mv^2}{r+h}$

Magnetic force,F=$qvB$

$\frac{mv^2}{r+h}=qvB$

$\frac{mv}{r+h}=qB$

Substitute the values

$\frac{1.67\times 10^{-27}\times (2.7\times 10^7)}{6.38\times 10^6+h}=1.6\times 10^{-19}\times 4\times 10^{-8}$

$6.38\times10^6+h=\frac{1.67\times 10^{-27}\times 2.7\times 10^7}{1.6\times 10^{-19}\times 4\times 10^{-8}}$

$6.38\times10^6+h=7.045\times 10^6$

$h=7.045\times 10^6-6.38\times 10^6$

$h=0.665\times 10^6=6.65\times 10^5 m$