Your employer asks you to build a 24-cm long solenoid with an interior field of 4.6 mT. The specifications call for a single layer of wire, wound with the coils as close together as possible. You have two spools of wire available. Wire with a #18 gauge has a diameter of 1.02 mm and has a maximum current rating of 6 A. Wire with a #26 gauge is 0.41 mm in diameter and can carry up to 1 A.

a. Which wire should you use?

b. What current will you need?

Explanation:

Formula for magnetic field in a solenoid

B = μ₀ N/ L x I

N is total number of turns , L is length of solenoid , I is current , B is magnetic field ,  μ₀ is constant equal to 4π x 10⁻⁷

Putting the values for 18 gauge wire

N = length of wire / diameter of wire

N = 24 / .102

= 235.3

B = μ₀ N/ L x I

4.6 x 10⁻³ = 4π x 10⁻⁷ x (235.3 / .24) x I

I = 3.735 A

Putting the values for 26 gauge wire

N = length of wire / diameter of wire

N = 24 / .041

= 585.36

B = μ₀ N/ L x I

4.6 x 10⁻³ = 4π x 10⁻⁷ x (585.36 / .24) x I

I = 1.5  A

a ) So 18 gauge wire will be used

b ) as current required in this case is

3.735 A

In the second case current required is 1.5 A which is beyond its tolerence limit.

float.

archimedes' principle

denser than water sinks in water.

denser than mercury would sink in it

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft