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

Answer :

Answer:

[tex]B=0.27840\ T[/tex]

Explanation:

Given:

mass of the wire, [tex]m=0.05\ kg[/tex]

length of the wire, [tex]L=0.16\ m[/tex]

current in the wire, [tex]I=11\ A[/tex]

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:

[tex]F=I.L\times B[/tex]

[tex]m.g=I.L\times B[/tex]

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

[tex]B=[/tex] magnetic field

[tex]0.05\times 9.8=11\times 0.16\times B[/tex]

[tex]B=0.27840\ T[/tex]

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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