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A wire carrying a 27.5 A current passes between the poles of a strong magnet perpendicular to its field and experiences a 2.19 N force on the 4.00 cm of wire in the field. What is the average field strength (in T)

Respuesta :

Answer:

B=1.99 T  

Explanation:

Given that

Current ,I = 27.5 A

Force ,F= 2.19 N

Length ,L = 4 cm = 0.0 4 m

We know that force on a magnet is given as

F= I B L sin θ  

B=Magnetic filed

F=Force

L=Length  

I=Current  

Given that current and force is perpendicular to each other that is why

     θ = 90 °  

2.19 = 27.5 x 0.04 x B x sin 90 °               ( sin 90 °  = 1 )

[tex]B=\dfrac{2.19}{27.5\times 0.04}\ T[/tex]

B=1.99 T

Therefore the value of magnetic field will be 1.99 T.

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