Question

At a certain location, the horizontal component of the earth's magnetic field is 2.7 × 10-5 T, due north. A proton moves eastward with just the right speed, so the magnetic force on it balances its weight. Find the speed of the proton.

Answer 1

To find the speed of the proton, we equate the gravitational force (mg) with the magnetic force exerted on a charge moving in a magnetic field (qvB). Given the proton's charge and mass, the known magnetic field, and Earth's gravitational acceleration, we can calculate the proton's speed using the equation v = mg / (qB).

Step-by-step explanation:

To calculate the speed of the proton that has a magnetic force balancing its weight, we can use the relationship between the magnetic force and the magnetic field. The magnetic force, F, on a moving charge, q, in a magnetic field, B, is given by F = qvB sin(θ), where v is the velocity of the charge, B is the magnetic field's strength, and θ is the angle between the velocity and the magnetic field.

In the student's case, the weight of the proton is balanced by the magnetic force. We can equate the magnetic force to the gravitational force on the proton, mg. This gravitational force is mg = qvB since the movement of the proton is perpendicular to the direction of the magnetic field (θ = 90 degrees, and sin(90) = 1). Solving for v, we have v = mg / (qB).

The charge of a proton (q) is approximately 1.602 x 10-19 C, the mass of a proton (m) is approximately 1.672 x 10-27 kg, and g is the acceleration due to gravity, approximately 9.81 m/s2. The magnetic field strength B is given as 2.7 x 10-5 T. Inserting these values into the formula gives the speed, v, of the proton.

Answer 2

Therefore the velocity of the proton is
`3.79 * 10^(-3)` m/s towards east.

Step-by-step explanation:

The mass of proton is m=
`1.67* 10^(-27)` kg.

q=
`1.6* 10^(-19)` C is the magnitude of the charge of a proton.

The earth's magnetic field is B=
`2.7* 10^(-5)` T toward north.

Let v be the velocity of the proton towards east.

The angle between the velocity of proton towards east and magnetic field towards north is
`90^\circ`.

We know that,

`F=|q|Bvsin \theta`

`\Rightarrow mg=|q|Bvsin \theta`

`\Rightarrow v=(mg)/(|q|Bsin\theta)`

Now putting the all values

`\therefore v=(1.67* 10^(-27)* 9.8)/(1.6* 10^(-19)* 2.7* 10^(-5) sin 90^\circ)`

`=3.79 * 10^(-3)` m/s

Therefore the velocity of the proton is
`3.79 * 10^(-3)` m/s towards east.