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A Voice Coil. The net force on a current loop in a uniform magnetic field is zero. The magnetic force on the voice coil of a loudspeaker see in figure is nonzero because the magnetic field at the coil is not uniform. A voice coil in a loudspeaker has 50 turns of wire and a diameter of 1.56 cm, and the current in the coil is . Assume that the magnetic field at each point of the coil has a constant magnitude of 0.220 T and is directed at an angle of $\mathbf{60}\mathbf{掳}$outward from the normal to the plane of the coil. Let the axis of the coil be in the y-direction. The current in the coil is in the direction shown (counterclockwise as viewed from a point above the coil on the y-axis). Calculate the magnitude and direction of the net magnetic force on the coil.

Step by step solution

01

Definition of magnetic field

The term magnetic field may be defined as the area around the magnet behave like a magnet.

02

Determine the magnitude and direction of the net magnetic force on the coil.

The force acting on segment dF is given by dF = ldlB the x component of the force cancel and the y component of force is

$\begin{array}{r}{\mathrm{dF}}_{\mathrm{y}}=\mathrm{dFcos}鈦�\left({30}^{鈭�}\right)\\ {\mathrm{dF}}_{\mathrm{y}}=鈭�\mathrm{Bcos}鈦�\left({30}^{鈭�}\right)\end{array}$

Integration on both sides

$\begin{array}{r}\mathrm{F}=鈭�{\mathrm{dF}}_{\mathrm{y}}\\ \mathrm{F}=|\mathrm{Bcos}鈦�\left({30}^{鈭�}\right)鈭�\mathrm{d}|\end{array}$

The integration of is is the total length of the coil

So

$\begin{array}{r}\mathrm{F}=\mathrm{IBcos}鈦�\left({30}^{鈭�}\right)\mathrm{N}\left(2\mathrm{蟺谤}\right)\\ \mathrm{F}=(0.950\mathrm{A})(0.220\mathrm{T})\left(\cos 鈦�\left(30{.0}^{鈭�}\right)\right)\left(50\right(2\mathrm{蟺}\left)\right)(0.0078\mathrm{m})\\ \mathrm{F}=0.444\mathrm{N}\end{array}$

And direction from diagram

$\stackrel{鈫�}{F}=\left(0.444N\right)\widehat{j}$

Hence, the magnitude and direction of the net magnetic force on the coil is .

$\stackrel{鈫�}{\mathrm{F}}=-(0.444\mathrm{N})\widehat{\mathrm{j}}$

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