what is the acceleration of a proton moving with a speed of 7.5m/s at right angles to a magnetic field of 1.4T?
To find the acceleration of a charged particle moving in a magnetic field, you can use the equation:
acceleration = (charge * velocity * magnetic field) / mass
In this case, we are given the following parameters:
- Charge of a proton (q) = +1.6 x 10^-19 C (Coulombs)
- Velocity of the proton (v) = 7.5 m/s
- Magnetic field (B) = 1.4 Tesla (T)
- Mass of a proton (m) = 1.67 x 10^-27 kg
Plugging in these values into the equation, we get:
acceleration = (1.6 x 10^-19 C * 7.5 m/s * 1.4 T) / 1.67 x 10^-27 kg
Now, let's calculate the acceleration.
To find the acceleration of a proton moving at right angles to a magnetic field, you can use the following equation:
F = qvB
Where:
- F is the force experienced by the proton
- q is the charge of the proton (1.6 x 10^-19 C)
- v is the velocity of the proton (7.5 m/s)
- B is the magnetic field strength (1.4 T)
In this case, the force experienced by the proton will act as the centripetal force, causing the proton to move in a circular path. The acceleration can then be calculated using the equation:
F = ma
Since we want to find the acceleration, we can rearrange the first equation to solve for a:
a = F/m
First, let's calculate the force experienced by the proton:
F = (q)(v)(B)
= (1.6 x 10^-19 C)(7.5 m/s)(1.4 T)
Now, we can use the mass of a proton (m = 1.67 x 10^-27 kg) to find the acceleration:
a = F/m
= [(1.6 x 10^-19 C)(7.5 m/s)(1.4 T)] / (1.67 x 10^-27 kg)
Simplifying this calculation will give you the acceleration of a proton moving at right angles to a magnetic field.