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aksik [14]
3 years ago
7

Suppose a proton ( = 1. 67×10^−27 kg) is confined to a box of width = 1. 00×10^−14 m (a typical nuclear radius).

Physics
1 answer:
dimulka [17.4K]3 years ago
3 0

Answer:

22e837281949222324

Explanation:

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1) Re-arranging the previous formula, we can calculate the value of the magnetic field intensity. The charge is q=2.7 \mu C=2.7 \cdot 10^{-6}C. In this case, v and B are perpendicular, so \theta=90^{\circ}, therefore we have:
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