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A 15.0 cm diameter circular loop of wire is placed with the plane of the loop parallel to the uniform magnetic field between the pole pieces of a large magnet. When 5.30 A flows in the coil, the torque on it is 0.135 m⋅N. What is the magnetic field strength please?

Respuesta :

Samueloladosu37

Answer:0.0144 tesla

Explanation:

Magnetic field strength is the same as Magnetomotive force

B= I/2πr

B=magnetic field strength

I=current

r=radius=0.15/2=0.75m

Area=πr2= 1.77m2

torqueT= NIABsin theta

B= T/NIABsin theta

B=0.135/1 * 5.3 * 1.77 *sin90

B=0.0144 tesla

Parrain

Based on the diameter of the loop, the current flowing, and the torque, the magnetic field strength will be 1.42T.

What is the magnetic field?

This can be found as:

= Torque / ( Current x Area x No. of coils)

Area is:

= π / 4 x diameter in meters ²

= π / 4 x 0.15²

= 0.018

Magnetic field strength is:

= 0.135 / (5.3 x 0.018 x 1)

= 1.42 T

Find out more on magnetic fields at https://brainly.com/question/26257705.

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