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A proton moves with a speed of 4.00 106 m/s horizontally, at a right angle to a magnetic field. What magnetic field strength is required to just balance the weight of the proton and keep it moving horizontally? (The mass and charge of the proton are 1.67 ✕ 10−27 kg and 1.60 ✕ 10−19 C, respectively.)

User Dan Def
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Answer:

Magnetic field,
B=2.55* 10^(-14)\ T

Step-by-step explanation:

It is given that,

Speed of proton,
v=4* 10^6\ m/s

Mass of the proton,
m=1.67* 10^(-27)\ kg

Charge on proton,
q=1.6* 10^(-19)\ C

We need to find the magnetic field strength required to just balance the weight of the proton and keep it moving horizontally.

The Lorentz force is given by :


F=q(v* B)=qvB\ sin90.............(1)

The weight of proton,


W=mg..............(2)

From equation (1) and (2), we get :


mg=qvB


B=(mg)/(qv)


B=(1.67* 10^(-27)\ kg* 9.8\ m/s^2)/(1.6* 10^(-19)\ C* 4* 10^6\ m/s)


B=2.55* 10^(-14)\ T

Hence, this is the required solution.

User Stack Player
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