Question

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.)

Answer 1

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.