Taking the radius of the earth as 6400 km and the horizontal component of the earth's magnetic field at the equator as 0.4×10−4 Wb/m2, the minimum momentum of a proton circling the earth at the equator is x×10−17 kg-m/s. The value of x is ______.[ Round off to the nearest integer ]

Taking the radius of the earth as 6400 km and the horizontal component

1.	Taking the radius of the earth as 6400 km and the horizontal component of the earth's magnetic field at the equator as 0.4×10−4 Wb/m2, the minimum momentum of a proton circling the earth at the equator is x×10−17 kg-m/s. The value of x is ______.[ Round off to the nearest integer ]
Answer» Taking the radius of the earth as $6400 km$ and the horizontal component of the earth's magnetic field at the equator as $0.4 \times 10^{- 4} {Wb/m}^{2}$ , the minimum momentum of a proton circling the earth at the equator is $x \times 10^{- 17} kg-m/s$ . The value of $x$ is ______. $[$ Round off to the nearest integer $]$

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Taking the radius of the earth as 6400 km and the horizontal component of the earth's magnetic field at the equator as 0.4×10−4 Wb/m2, the minimum momentum of a proton circling the earth at the equator is x×10−17 kg-m/s. The value of x is ______.[ Round off to the nearest integer ]

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