# An investigator places a sample 1.0 cm from a wire carrying a large current; the strength of the magnetic field has a particular value at this point. Later, she must move the sample to a 11.0 cm distance, but she would like to keep the field the same. Part A) By what factor must she increase the current?

Answer: she will have to increase the factor of current by 11

Explanation: The mathematical relationship between the strength of the magnetic field (B) created by a current carrying conductor with current (I) is given by the Bio-Savart law given below

B=

B=strength of magnetic field

I = current on conductor

r = distance on any point of the conductor from it center

u = permeability of magnetic field in space

from the question, the investigator is trying to keep a constant magnetic field meaning B has a fixed value such as the constants in the formulae, the only variables here are current (I) and distance (r). We can get this a mathematical function.

by cross multipying, we have

B* 2πr=I

by dividing through to make I subject of formulae, we have that

I =

B, 2π and are all constants, thus

= k(constant)

thus we have that

I =kr (current is proportional to distance assuming magnetic field strength and other parameters are constant)

thus we have that

=

=1cm and =11cm

=

thus =11*

which means the second current is 11 times the first current

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A 0.6 kg block attached to a spring of force constant 13.6 N/m oscillates with an amplitude of 9 cm. Find the maximum speed of the block. Answer in units of m/s. 003 (part 2 of 4) 10.0 points Find the speed of the block when it is 4.5 cm from the equilibrium position. Answer in units of m/s. 004 (part 3 of 4) 10.0 points Find its acceleration at 4.5 cm from the equilibrium position. Answer in units of m/s 2 . 005 (part 4 of 4) 10.0 points Find the time it takes the block to move from x = 0 to x = 4.5 cm. Answer in units of s.

1) 0.43 meters per second

2) 0.21 meters per second

3) 1.02

4) 0.66 seconds

Explanation:

part 1

By conservation of energy, the maximum kinetic energy (K) of the block is at equilibrium point where the potential energy is zero. So, at the equilibrium kinetic energy is equal to maximum potential energy (U):

With m the mass, v the speed, k the spring constant and xmax the maximum position respect equilibrium position. Solving for v

part 2

Again by conservation of energy we have kinetic energy equal potential energy:

part 3

Acceleration can be find using Newton's second law:

with F the force, m the mass and a the acceleration, but elastic force is -kx, so:

part 4

The period of an oscillator is the time it takes going from one extreme to the other one, that is going form 4.5 cm to -4.5 cm respect the equilibrium position. That period is:

So between 0 and 4.5 cm we have half a period:

A flock of geese is flying south for the winter. In order to maintain a “V” shape, each goose is flying at the same constant velocity. Which goose will have the most momentum? A) the goose with the least mass

B) the goose with the most mass

C) the goose that is behind all the others

D) the goose at the front of the "V" leading the way of the other geese