The magnetic field inside a solenoid with length Lo, radius Ro, number of turns No, and current Io is Bo. If the length and radius were doubled while keeping the number of turns and current the same, the magnetic field would be: -

If the length and radius were doubled keeping the number of turns and current the same, the magnetic field would be B0/2.

Explanation:

A solenoid is  a device that has wires wound in the shape of  a cylinder. It is used to generate strong magnetic field by passing current through it. The magnetic field inside a solenoid is determined by the current passing through the solenoid as well as the number of turns per unit length of the solenoid.

The diameter or radius of the solenoid has no effect on the magnetic field inside a solenoid.

B=μ0nI

where μ0 represents the permeability of free space, n represents the number of turns per unit length and I represents the current passing through the solenoid.

n is given by the expression

n=N/L where N is the total number of turns of the solenoid and L is its length

B0=μ0NI/L

in this problem

new magnetic field Bnew=μ0NI/2L0=

Hence the magnetic field will become half of its initial value.

c.the charges if the proton and electron and gravity

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer