A wire with mass per unit length of 1.00 g/cm is placed on a horizontal surface.

Question

A wire with mass per unit length of 1.00 g/cm is placed on a horizontal surface. the wire carries a current of 1.5 A eastward & moves horizontally to the north. If a magnetic field of .2 T points into the surface, find the surface's coefficient of friction of the wire moves with an acceleration of 1.04 m/s2.

Explanation / Answer

Hello!! Consider a right handed coordinate system (CS) with the (+x)-axis pointing eastward, the (+y)-axis pointing northward. With the current carrying wire lying along the x-axis and moving northward in the (+y)-direction due to the force exerted by the vertical magnetic field B, then according to the right hand rule of vector cross product, B must be pointing downward along the (-z)-direction. The magnitude of the magnetic force is (1) ..... F = i*L*B*sin90° = i*L*B The magnetic force must overcome the static friction (2) ..... Fs = µ*N = µ*M*g in order to make the wire move. That is, (3) ..... F = i*L*B >= µ*M*g The equality gives the minimum B that will make the wire move. Solving for B, we get (4) ..... B = µ*M*g / (i*L) = µ* (M/L)*g / i The quantity M/L is just the given mass per unit length of the wire that is equal to 1.00 g/cm. Substituting the values, we get the minimum magnetic field B = 1.3 tesla Hope this helps! Have anice day!!

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