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A rectangular solid conducting bar having a square cross-section of side \"a\" a

ID: 2019204 • Letter: A

Question

A rectangular solid conducting bar having a square cross-section of side "a" and length L is placed near the surface of the Earth at the top of a pait of smooth parallel superconducting rails forming a ramp angle . THe bar is made of frictionless material having a mass per unit volume (density) D and a resistivity . The rails are separated by the distance L and interconnected by a superconducting section at the bottom such that a conducting loop is created when the bar is positioned on the rails. A verical magnetic filed B having a constant direction along -j but changing magnitude exists between section and the bar remains at rest. g is the freefall rate of acceleration and t is the time such that |B| = B0 when t = 0...

Determine an experssion for B in terms of t, S, , g, D and

Explanation / Answer

The bar will keep accelerating until the force from the current generated cancels the force of gravity. The force of gravity on the bar pointing down the rails is             F = IBL The component of magnetic field normal to the block                                    B = Bcos    F   =   IBLcos The component of gravitational force acting on the bar mgsin   =  IB0Lcos ==>  B0   =   mgtan / IL    density of the rod   D = m   / aL      ==> m = mass of the rod = DaL    I   =   DaLgtan / B0L     ==>   Dagtan    / B0   ..........(1) The induced emf produced along the length of the bar                             = change in magnetic flux = d /dt =   A(dB /dt)              Induced emf   = IR , A = change in magnetic flux threough an area = sL         Here I = induced current , R = resistance of the bar = L / a                I (L / a) =  A(dB /dt)                          (dB /dt)      =    I(L / asL)                   dB   =   I(L / asL) dt intergrating on both sides we get                          B(t) =     I(L / asL)t          From equation (1) I = Dagtan    / B0     Hnce    B (t) = (Dagtan    / B0 ) (L / asL)t      = = (Dagtan    / B0 ) ( / as)t                                                     B = Bcos    F   =   IBLcos The component of gravitational force acting on the bar mgsin   =  IB0Lcos ==>  B0   =   mgtan / IL    density of the rod   D = m   / aL      ==> m = mass of the rod = DaL    I   =   DaLgtan / B0L     ==>   Dagtan    / B0   ..........(1) The induced emf produced along the length of the bar                             = change in magnetic flux = d /dt =   A(dB /dt)              Induced emf   = IR , A = change in magnetic flux threough an area = sL         Here I = induced current , R = resistance of the bar = L / a                I (L / a) =  A(dB /dt)                          (dB /dt)      =    I(L / asL)                   dB   =   I(L / asL) dt intergrating on both sides we get                          B(t) =     I(L / asL)t          From equation (1) I = Dagtan    / B0     Hnce    B (t) = (Dagtan    / B0 ) (L / asL)t      = = (Dagtan    / B0 ) ( / as)t                 
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