Question

In cosmic rays, 0.15 proton cm^-2 s^-1 are entering the
earth's atmosphere. If the earth's radius is 6400 km, the
current received by the earth in the form of cosmic rays is
nearly equal to
(a) 0.12A (b) 1.2A (C) 12A (d) 120 A.​

Answer 1

Answer: a) 0.12 A

Explanation: Let Q denote the amount of charge entering the Earth's atmosphere per unit area per unit time.

Now, charge on each proton = 1.6 X 10^(-19) C

Therefore,

Q = 1.6 X 10^(-19) X 0.15 C-cm^(-2)-s^(-1)

I'll switch cm^(-2) to m^(-2) for convenience. Applying the conversion factor seems correctly is crucial to this problem, as is visible from the solutions.

Q = 1.6 X 10^(-19) X 0.15 X 10^(4) C-m^(-2)-s^(-1)

Q = 2.4 X 10^(-16)  C-m^(-2)-s^(-1)

Now, the current passing through the entire atmosphere will be Q times the surface area of Earth's atmosphere.

I = A x Q x t

t = 1 s, so I'll simply drop the time unit from Q.

I = 4 X π X R^2 X Q A(Here, A denotes Ampere)

I = 4 X 3.14 X (6.4 x 10^6)^2 X 2.4 X 10^(-16) A

I = 0.12 A, as you can now easily verify.

Answer 2

Answer:

The current received by the earth is (a) 0.12 A

Explanation:

Cosmic rays are the protons and nuclei that travels approximately with the speed of light prevading in the space. Here, cosmic rays striking the earth surface and forming a current.

Cosmic rays are striking the earth surface per unit area per unit second. It is given that protons are striking the earth surface. The charge q that is coming from the cosmic rays per unit area per unit second is calculated from the given value of

$q = 0.15×charge\:on\:one\:proton\:cm^{-2} s^{-1}$

$q = 0.15×1.6×10^{-19} Ccm^{-2}s^{-1}$

Converting the above in SI units,

$q = 0.15×1.6×10^{-19}×10^{4} Cm^{-2}s^{-1}$

$q = 2.4×10^{-16} Cm^{-2}s^{-1}$

Now, calculate the total surface area of the earth.

The radius of the earth = 6400 km = $6.4×10^{6} m$

Area of earth =$4×π×(radius)^{2}$

= $4×3.14×(6.4×10^{6})^{2} m^{2}$

= $514×10^{12} m^{2}$

= $5.14×10^{14} m^{2}$

The current received by the earth = Area of earth×q

= $5.14×10^{14}×2.4×10^{-16} C/s$

= $0.12 A$

So, the current received by the earth = 0.12 A