Figure 31N-1 \\a shows a rectangular conducting loop of resistance R = 0.020 ohm

Question

Figure 31N-1  shows a rectangular conducting loop of resistance R = 0.020 ohm, height H = 1.5 cm. and length D = 2.5 cm being pulled at constant speed v = 40 cm/s through two regions of uniform magnetic field. Figure 31N-11b gives the current i induced in the loop as a function of the position x of the right side of the loop. For example, a current of 3.0 mu A is induced clockwise as the loop enters region 1. What are the magnitudes and directions of the magnetic field in region 1 and region 2?

Explanation / Answer

According to faraday 's law induced emf    = d_B / dt Here _B = magnetic flux linked with the recangular loop let is suppose that in region 2 the length of the loop is x the remaining D - x potrion in region (1) hence _B = xHB_2 + (D-x) HB_1 = DHB_1 + xH(B2 - B1) d_B / dt    = (dx/dt )*H(B2 - B1)  = v*H(B2 - B1) Hence indice current I = v*H(B2 - B1) / R Here R = resistance of the loop From fig (b) for region (1) initially B_1 = 0.0 T and   B_2   = B_1 I    = 3.0*10^-6 A = (0.4 m /s) *0.015m (B_1 - 0 ) / 0.02 Hence solving For B_1 we get = 10 T According to lenz's law The direction of B_1 is out of the page similarly soving For B_2 I =  v*H(B2 - B1) / R    Here B_1 = 10 T , Indiced current I = 2.0*10^-6 T , v = 0.4 m/s , H = 0.015m we get B_2 = 3.3 T According to Lenz's Law direction of magnetic filed   is out of the page

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