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What is the value of B (in ×10-8 T ) that can be set up at the equator to permit a proton of speed 107 m/s to circulate around the earth? R=6.4×106 m,mP=1.67×10-27 kg

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answer is 1.6.

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Detailed Solution

From the equation, r=mvBq  We have B=mvqr Substituting the values, we have B=1.67×10-271071.6×10-196.4×106=1.6×10-8 T
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