A wire lying along a y axis from y=0 to y=0.250 m carries a current of 2.00 mA in the negative d

A wire lying along a y axis from y=0 to y=0.250 m carries a...

A wire lying along a $y$ axis from $y=0$ to $y=0.250 \mathrm{m}$ carries a current of 2.00 $\mathrm{mA}$ in the negative direction of the axis. The wire fully lies in a nonuniform magnetic field that is given by $B=(0.300 \mathrm{T} / \mathrm{m}) y \hat{\mathrm{i}}+(0.400 \mathrm{T} / \mathrm{m})$ yj. In unit-vector notation, what is the magnetic force on the wire?

A wire lying along a $y$ axis from $y=0$ to $y=0.250 \mathrm{m}$ carries a current of 2.00 $\mathrm{mA}$ in the negative direction of the axis. The wire
fully lies in a nonuniform magnetic field that is given by
$B=(0.300 \mathrm{T} / \mathrm{m}) y \hat{\mathrm{i}}+(0.400 \mathrm{T} / \mathrm{m})$ yj. In unit-vector notation, what is the magnetic force on the wire?

Key Concepts

Nonuniform Magnetic Fields

In many practical situations, the magnetic field is not constant in space; its magnitude or direction may vary with position. This nonuniformity requires careful integration of the Lorentz force over the wire, since the magnetic field B must be evaluated at each differential element of the wire. Understanding how to incorporate spatially-varying magnetic fields into the force calculation is essential in accurately predicting the net effect on the current-carrying conductor.

Cross Product and Right-Hand Rule

The cross product in vector calculus is used to calculate the force direction on a current element within a magnetic field. In the context of the Lorentz force, taking the cross product of the current element vector with the magnetic field vector yields a force vector that is perpendicular to both. The right-hand rule is a mnemonic for determining the direction of the resultant vector, ensuring the proper representation of the force's direction in space.

Lorentz Force

This concept refers to the force experienced by a current-carrying conductor in a magnetic field. The force is given by the differential form dF = I (d? × B), where I is the current, d? is a differential length element of the wire, and B is the magnetic field. This relationship is fundamental in electromagnetism and is applied by integrating over the entire length of the conductor to find the net force.

Differential Elements and Integration

When dealing with a current-carrying wire, the force is found by breaking the wire into infinitesimal segments, each experiencing a differential force. By expressing the position and the current direction in terms of differential elements and then integrating over the wire’s length, one accounts for variations in both current direction and magnetic field at different points along the wire.

Transcript

00:01 Here a wire is lying along x -axis from y equal to 0 to y is equal to 0 .25 and current is carried by the wire in negative direction of the axis so minus g direction so that is current is 0 .00 millie -amper and that equals to 2 into 10 to the power minus 3 ampere and that is in g -hidreakricon so let me that that has minus g hat all right and this is an emperor now now what else is given that there is a uniform magnetic field b equals to okay so the magnetic field is dependent upon is varying along y axis so it says it'll be easier if i don't write zero the value is the same so tesla per meter and y it is dependent upon y it is varying the magnetic field is varying along y axis plus 0 .4 tesla per meter along y and j hat all right so we are asked to calculate the magnetic force on the wire f is equal to what so f is equal to i d l it is the cross product of i d l and b right we consider that the direction of length and current is the same so ideal is as a whole of vector quantity current and the length so if you multiply i i i hat and b so here if you see here this component j cross j will turn out to be 0 and we are only so this let me write this as small force df okay because we have considered a small length we are considering force on a small length deal so let me write this as df not so let me get back again here so the current here if it is i cross b then that will give me and j cross j will be so the only term left will be this this cross this right so we'll to that df equals to what is this here minus 2 into 10 to the power minus 3 right and what is b the constant in b is 0 .3 and we are left with now y and dl is dl.

03:25 Let me write this dl as d .y this will make it easier to integrate right so let us change this formula as d -y because the length is along y -exists all right now the length goes from 0 to 0 .25 right so let us integrate this to find out the total force f will be into sum of all forces from length 0 to 0 .25 all right so this will give me 0 to 0 .0...

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