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ohaa [14]
2 years ago
7

Assuming the same current is running through two separate coils, why is it easier to thrust a magnet into a wire coil with one l

oop than one with four loops?
Physics
1 answer:
alexira [117]2 years ago
3 0

it is easier to thrust a magnet into a wire coil with one loop than one with four loops Because with four loops it becomes more difficult to push the magnet in.

<h3>What is a magnetic field?</h3>

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

The quantity of turns or loops in a coil directly relates to the intensity of the magnetic field within the coil.

As a result, the magnetic field strength has increased fourfold when there are four loops instead of one, making it more difficult to push the magnet in.

A single-looped wire coil is simpler to insert a magnet into than one with four loops. because inserting the magnet gets more challenging when there are four loops.

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

brainly.com/question/19542022

#SPJ1

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At what time was the person at a position of 0m?
Anni [7]

Answer: The person was not at a position of "0" at any time. The person started at 10 metres from the starting line. The explanation below shows how to use the standard formula for position when the initial position is not "0". It is noteworthy that the standard expression of the formula for distance travelled does not include a variable (e.g. "d") for distance at the start (when t(time) = 0)

Explanation: At time = 0, the start, the person was at 10m distance from the starting line. Therefore, to use the standard equation, "s + ut + 1/2att (t squared, that is), distance from starting line = 10 + s, that is, total distance from starting line  equals initial position, 10 metres, plus "s" (distance travelled from t = 0 to t = 1) in metres.

for the section of the graph from "0" seconds (t = 0) to 1 second (t = 1):

s = ut + 1/2att

the initial position is 10 metres.

s = 10

the distance is constant from t = 0 to t = 1, therefore the velocity for the whole of that section of graph must be 0.

u = 0

there is no change in the velocity from t = 0 to t= 1, therefore the acceleration for the first section of the graph must be 0.

a = 0

s = ut + 1/2att

  = (0 x 1) + 1/2 (0 x 1 x 1)

  = (0) + 1/2 (0)

  = 0

total distance from starting line (position) equals initial position plus change in position (distance travelled).

at t = 1,

position = 10 + 0

 = 10 metres

The whole of the graph can be analysed using this process for each straight section of the graph separately, adding "s" for each section to the previous total of distance from starting line.

using "d" for initial distance from starting line ( position ), d1 for distance from starting line at t = 1, d2 for distance from starting line at t = 2, etcetera:

section 1, t = 0 to t = 1:

d1 (t=0 to t=1)  =  10 + s (t=0 to t=1).

section 2, t= 1 to t = 2:

d2 (t=0 to t=2) = 10 + s (t=0 to t=1) + s (t=1 to t=2).

etcetera.

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