A long, straight, horizontal wire carries a left-to-right current of 20 A. If the wire is placed in a uniform magnetic field of
1 answer:
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In step 1, to increase the potential energy, the iron will move towards the electromagnet.
In step 2, to increase the potential energy, the iron will move towards the electromagnet.<h3>Potential energy of a system of magnetic dipole</h3>The potential energy of a system of dipole depends on the orientation of the dipole in the magnetic field.
where;
is the dipole moment
- B is the magnetic field
Increase in the distance (r) reduces the potential energy. Thus, we can conclude the following;
- In step 1, to increase the potential energy, the iron will move towards the electromagnet.
- In step 2, when the iron is rotated 180, it will still maintain the original position, to increase the potential energy, the iron will move towards the electromagnet.
Learn more about potential energy in magnetic field here: brainly.com/question/14383738
Answer:
life (N) of the specimen is 117000 cycles
Explanation:
given data
ultimate strength Su = 120 kpsi
stress amplitude σa = 70 kpsi
solution
we first calculate the endurance limit of specimen Se i.e
Se = 0.5× Su .............1
Se = 0.5 × 120
Se = 60 kpsi
and we know strength of friction f = 0.82
and we take endurance limit Se is = 60 kpsi
so here coefficient value (a) will be
a = ......................1
put here value and we get
a =
a = 161.4 kpsi
so coefficient value (b) will be
b =
b =
b = −0.0716
so here number of cycle N will be
N =
put here value and we get
N =
N = 117000
so life (N) of the specimen is 117000 cycles