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

Answer:

Magnetic field, $B=2.55\times 10^{-14}\ T$

Explanation:

It is given that,

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

Mass of the proton, $m=1.67\times 10^{-27}\ kg$

Charge on proton, $q=1.6\times 10^{-19}\ C$

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 :

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

The weight of proton,

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

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

$mg=qvB$

$B=\dfrac{mg}{qv}$

$B=\dfrac{1.67\times 10^{-27}\ kg\times 9.8\ m/s^2}{1.6\times 10^{-19}\ C\times 4\times 10^6\ m/s}$

$B=2.55\times 10^{-14}\ T$

Hence, this is the required solution.