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3 years ago

16

A magnetically soft material is placed in a strong magnetic field. What is the most likely outcome?

2 answers:

Svet_ta [14]3 years ago

9 0

D-It will become a temporary magnet because the domains will easily realign.

natka813 [3]3 years ago

5 0

Answer: The correct answer is D.

Explanation:

When the iron is placed in the magnetic field then it will magnetize and behave like a temporary magnet. The domains of the iron will align.

Magnetically soft material: It is an temporary magnet which is made from a material that is magnetized. It creates its own magnetic field. For example, magnetically soft material is an electromagnet. The direction of this material can be changed easily.

In the given problem, a magnetically soft material is placed in a strong magnetic field. The domains of the magnetically soft material are already aligned but when this material is placed in the magnetic field then the domains will easily realign.

Therefore, the magnetically soft material will become a temporary magnet because the domains will easily realign.

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Lera25 [3.4K]

Answer:

$a_{cA} = 6.075$ m/s²

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Where:

$a_{c}$ : Is the normal or centripetal acceleration of the body ( m/s²)

v: It is the magnitude of the tangential velocity of the body at the given point

.(m/s)

r: It is the radius of curvature. (m)

Newton's second law:

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∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

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$v_{B} = 27 \frac{m}{s}$

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Problem development

We replace data in formula (1) to calculate centripetal acceleration:

$a_{cA} = \frac{(27)^{2} }{120}$

$a_{cA} = 6.075$ m/s²

$a_{cB} = \frac{(27)^{2} }{120}$

$a_{cB} = 6.075$ m/s²

We replace data in formula (2) to calculate centripetal force Fc) :

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$F_{cB} = m_{B} *a_{cB} = 1600kg*6.075\frac{m}{s^{2} }$

$F_{cB} = 9720 N$

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