Question

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 1

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.