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Simora [160]
3 years ago
7

A magnetic field is used to suspend a wire of mass 0.05 kgkg and length 0.16 mm . The wire is carrying a current of 11 AA . What

minimum magnetic-field magnitude is needed to balance the pull of gravity
Physics
1 answer:
pochemuha3 years ago
7 0

Answer:

B=0.27840\ T

Explanation:

Given:

mass of the wire, m=0.05\ kg

length of the wire, L=0.16\ m

current in the wire, I=11\ A

Now form the Lorentz force we know that every current carrying wire experiences a magnetic force on it due to the magnetic field components perpendicular to it.

Mathematically given as:

F=I.L\times B

m.g=I.L\times B

here: we nee to balance the gravitational force on the mass.

B= magnetic field

0.05\times 9.8=11\times 0.16\times B

B=0.27840\ T

This magnetic field must be perpendicular to the direction of current in the wire and the these two directional quantities must lie in a plane normal  to the direction of gravity according to the Fleming's left hand rule.

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(a) The moment of inertia of the sign for rotation about the side of length h is 1.24 kgm².

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The moment of inertia of the sign for rotation about the side of length h is calculated as follows;

I = 2 x ¹/₃M(b/2)²

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Learn more about rotational kinetic energy here: brainly.com/question/25803184

#SPJ1

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