A straight piece of conducting wire with mass M and length L is placed on a friction less incline tilted at an angle u from the horizontal . There is a uniform, vertical magnetic field $\overrightarrow{\mathbf{B}}$at all points (produced by an arrangement of magnets not shown in the figure). To keep the wire from sliding down the incline, a voltage source is attached to the ends of the wire. When just the right amount of current flows through the wire, the wire remains at rest. Determine the magnitude and direction of the current in the wire that will cause the wire to remain at rest. Copy the figure and draw the direction of the current on your copy. In addition, show in a free-body diagram all the forces that act on the wire.

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 current in the wire that will cause the wire to remain at rest.

${\mathrm{F}}_1=\mathrm{ILB}$On the wire the magnetic force is calculate as

${\mathrm{F}}_1=\mathrm{ILBsin}\left(\mathrm{\varphi }\right)$

The value of$\varphi =90°$

The summation of forces in vertical direction (y-axis) is zero$\sum F_y=0$

Or

$F_1\mathrm{Cos}\left(\theta \right)-Mg\sin \left(\theta \right)=0$

Put above result

$\begin{array}{r}\text{ILBCos}\left(\theta \right)-\mathrm{Mg}\sin \left(\theta \right)=0\\ \text{I=}\frac{\mathrm{Mgsin}\left(\theta \right)}{\mathrm{LB}}\end{array}$

Hence, the magnitude and direction of the current in the wire that will cause the wire to remain at rest is $\mathrm{I}=\frac{\mathrm{Mgtan}\left(\theta \right)}{\mathrm{LB}}$.

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