Physics: A wire lying along a y axis form y=0 to y=0.25 m carries a current of 2.0 mA in the negative direction of the axis. The wire fully lies in a magnetic field given by (0.3y)i + (0.4y)j Tesla. What isThe magnetic force on the wire

A wire lying along a y axis form y=0 to y=0.25 m

Ответ:

Emorej22

27.01.2022

Physics

FB = 0.187 N

Explanation:

To find the magnetic force FB in the wire you use the following formula:

$|\vec{F_B}|=ILBsin\theta\\\\L=0.25m\\\\|\vec{B}|=\sqrt{(0.3y)^2+(0.4y)^2}=0.5y \ T$

the angle between B and L is given by:

$\vec{L}\cdot\vec{B}=LBcos\theta\\\\\theta=cos^{-1}(\frac{\vec{L}\cdot\vec{B}}{LB})=cos^{-1}(\frac{0*0.3y+0.25*0.4y}{0.25*0.5y})=36.86\°$

Due to B depends on "y" you take into account the contribution of each element dy of the wire to the magnitude of the magnetic force. Thus, you have to integrate the following expression:

$|\vec{F_B}|=Isin\theta\int_0^{0.25}B(y)dy=Isin\theta\int_0^{0.25}(0.5y)dy\\\\|\vec{F_B}|=(2.0*10^{-3}A)(sin36.86\°)(0.5T)[\frac{0.25^2}{2}m]=0.187\ N$

hence, the magnitude of the magnetic force is 0.187N

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Ответ:

Angel1107

22.08.2021

Physics

Hello There!!

Explanation:

Pure copper cannot be made in to a magnet because it doesn't have magnetic domains and also copper is weakly magnetic.

hope this helps,have a great day!!

~Pinky~

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A wire lying along a y axis form y=0 to y=0.25 m carries a current of 2.0 mA in the negative direction of the axis. The wire fully lies in a magnetic field given by (0.3y)i + (0.4y)j Tesla. What isThe magnetic force on the wire

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