21 A 6.75 g wire of length L=15.0 cm is suspended by a pair of flexible leads in a uniform mag

step 1 In this question, we have this setup. So we have a wire that is being supported by two springs.

21 A 6.75 g wire of length L=15.0 cm is suspended by a...

21 A $6.75 \mathrm{~g}$ wire of length $L=15.0 \mathrm{~cm}$ is suspended by a pair of flexible leads in a uniform magnetic field of magnitude $0.440 \mathrm{~T}$ (Fig. $28-30$ ). What are the (a) magnitude and (b) direction (left or right) of the current required to ing leads?

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Key Concepts

Magnetic Force on a Current-Carrying Conductor

This concept deals with the force experienced by a current-carrying wire when it is placed in a magnetic field. The force is given by the equation F = I L B sin(?), where I is the current, L is the length of the wire, B is the magnetic field strength, and ? is the angle between the wire and the magnetic field. In many problems, the wire is oriented such that sin(?) is 1 (i.e., perpendicular), simplifying the expression to F = I L B.

Force Equilibrium

In scenarios where a current-carrying wire is suspended, the magnetic force must counteract the gravitational force acting on the wire for equilibrium. This means that the magnetic force (I L B) must equal the weight of the wire (mg), allowing one to solve for the required current using the balance mg = I L B.

Right-Hand Rule

This is a mnemonic used to determine the direction of the magnetic force on a current-carrying conductor. By pointing the thumb of the right hand in the direction of the current and the fingers in the direction of the magnetic field, the force is indicated by the direction in which the palm pushes. This rule helps identify whether the current should flow left or right to produce the necessary force direction.

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