At a certain location, the horizontal component of the earth’s magnetic field is

Question

At a certain location, the horizontal component of the earth’s magnetic field is 2.5E-5 T, due north. A proton moves eastward with just the right speed so that the magnetic force on it balances its weight. Find the speed of the proton. (Proton mass =1.67 × 10-27Kg At a certain location, the horizontal component of the earth’s magnetic field is 2.5E-5 T, due north. A proton moves eastward with just the right speed so that the magnetic force on it balances its weight. Find the speed of the proton. (Proton mass =1.67 × 10-27Kg At a certain location, the horizontal component of the earth’s magnetic field is 2.5E-5 T, due north. A proton moves eastward with just the right speed so that the magnetic force on it balances its weight. Find the speed of the proton. (Proton mass =1.67 × 10-27Kg

Explanation / Answer

Apply, FB = Fg

q*v*B*sin(90) = m*g

v = m*g/(q*B)

= 1.67*10^-27*9.8/(1.6*10^-19*2.5*10^-5)

= 4.09*10^-3 m/s

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