A current-carrying wire of length 50 cm is positioned perpendicular to a uniform magnetic field. The magnetic field strength is 0.6 Tesla.
<h3>What is magnetic field?</h3>
The magnetic field is the region of space where an object experiences a magnetic force as it enters the field.
Given is the wire of length 50 cm = 0.5 m and the current is 10.0 A. There is a resultant force of 3.0 N on the wire due to the interaction of the current and field.
The relation between magnetic field strength and current is
F = ILB
Substituting 50 for L, 10 for I and 3 for F, we get the magnetic field strength B.
3 = 10 × 0.5 × B
B= 0.6 Tesla
Therefore, the magnetic field strength is 0.6 Tesla.
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Answer:
The coefficient of thermal expansion of material B is more than A
Explanation:
As we know that when we apply heat to a material it expands. The amount of expansion depends on the nature of the material. The amount of expansion depends on the coefficient of thermal expansion of the material.
If a material has high value of thermal expansion coefficient then it expands more.
here material A expands less and material B expands more because the pin which is made by material A becomes loosen, so the coefficient of thermal expansion of B is more than the coefficient of thermal expansion of A
Are there answer choices?
Answer:
the ball travelled approximately 60 m towards north before stopping
Explanation:
Given the data in the question;
First course : 
= 0.75 m/s², 
= 20 m, 
= 10 m/s
now, form the third equation of motion;
v² = u² + 2as
we substitute
² = (10)² + (2 × 0.75 × 20)
² = 100 + 30
² = 130
= √130
= 11.4 m/s
for the Second Course:
= 11.4 m/s, 
= -1.15 m/s², 
= 0
Also, form the third equation of motion;
v² = u² + 2as
we substitute
0² = (11.4)² + (2 × (-1.15) × 
)
0 = 129.96 - 2.3
2.3 = 129.96
= 129.96 / 2.3
= 56.5 m
so;
|d| = √( ² + ² )
we substitute
|d| = √( (20)² + (56.5)² )
|d| = √( 400 + 3192.25 )
|d| = √( 3592.25 )
|d| = 59.9 m ≈ 60 m
Therefore, the ball travelled approximately 60 m towards north before stopping
