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A proton moves with a speed of 3.60 106 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.) B = T

User Jenniffer
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1 Answer

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Answer:

The magnetic field strength is required
2.84*10^(-14)\ T

Step-by-step explanation:

Given that,

Speed of proton
v = 3.60*10^(6)\ m/s

Mass of proton
m_(p)=1.67*10^(-27)\ kg

Charge
q =1.60*10^(-19)\ C

When a proton moves horizontally, at a right angle to a magnetic field .

Then, the gravitational force balances the magnetic field


mg=Bqv\sin\theta


B = (mg)/(qv)

Here,
\theta = 90^(\circ)

Where, B = magnetic field

q = charge

v = speed

Put the value into the formula


B = (1.67*10^(-27)*9.8)/(1.60*10^(-19)*3.60*10^(6))


B = 2.84*10^(-14)\ T

Hence, The magnetic field strength is required
2.84*10^(-14)\ T

User Caterham
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