Question

What is the acceleration of a proton moving with a speed of 9.5 m/s at right angles to a magnetic field of 1.4 t ?

Answer 1

The magnetic force $F=qvB$ provides the centripetal force that keeps the proton in circular motion:
$qvB=m \frac{v^2}{r}$
where the second term is the centripetal force, which is the product between the mass of the proton m and the centripetal acceleration: $a_c = \frac{v^2}{r}$
If we divide by m both terms, we can calculate the centripetal acceleration of the proton:
$a_c = \frac{qvB}{m}= \frac{(1.6 \cdot 10^{-19} C)(9.5 m/s)(1.4 T)}{1.67 \cdot 10^{-27} kg} =1.27 \cdot 10^9 m/s^2$

Answer 2

The acceleration of a proton will be 1.27 ×10⁹m/s².Acceleration is the rate of the velocity change.

What is a magnetic field?

It is the type of field where the magnetic force is obtained. With the help of a magnetic field. The magnetic force is obtained it is the field felt around a moving electric charge.

The given data in the problem is;

a is the acceleration of a proton=?

v is the speed = 9.5 m/s

B is the magnetic field = 1.4 T

The megnetic force is equated to the centripetal force;

$qvB= m\frac{v^2}{r} \\\\ qvB= m\frac{v^2}{r} \\\\ a_c= \frac{qvB}{m} \\\\\ \ a_c= \frac{1.6 \times 10^{-19}\times 9.5 \times 1.4}{1.67\times 10^{-27}} \\\\\ a_c=1.27 \times 10^9 m/s^2$

Hence the acceleration of a proton will be 1.27 ×10⁹ m/s².

To learn more about the magnetic field refer to the link;

https://brainly.com/question/19542022