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A telephone cable at a place has four long straight horizontal wires carrying a current of 1.0 A in the same direction east to west. The earth's magnetic field at the place is 0.39 G, and the angle of dip is 35^@. The magnetic declination is nearly zero. What are the resultant magnetic fields at points 4.0 cm below the cable ? |
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Answer» Solution :Here `N= 4 , I= 1.0 A, B_E = 0.39 G = 3.9 xx 10^(-5) T and ` angle of dip `delta=35^@` `therefore B_H= B_E cos delta = 3.9xx10^(-5) xx cos 35^@ = 3.9xx10^(-5) xx 0.8192 = 3.19 xx 10^(-5) T` and `B_V= B_V = B_E sin delta = 3.9xx10^(-5) xx sin 35^@ xx 0.5736 =2.24 xx10^(-5) T` and magnetic field due to current flowing in cables at a point situated at a distance R= 4 cm = 0.04 m is `B= (mu_0 N I)/(2pi r) =(4pi xx 10^(-7) xx 4 xx 1.0)/(2pi xx 0.04) = 2 xx 10^(-5) T` At a point 4 cm below the cable the cable applying right hand rule we find that B is in a direction opposite to that of `B_H` . Hence , horizontal component of resultant magnetic field. `B_(HR) = B_H - B = 3.19 xx 10^(-5) - 2 xx 10^(-5) = 1.19xx 10^(-5) T` and VERTICAL component of resutlant magnetic field `B_(VR) = B_V = 2.24 xx 10^(-5) T` `therefore ` Resultant magnetic field `B_g = sqrt((B_(HR))^2 + (B_(VR))^2) = sqrt((1.19xx10^(-5))^2 + (2.24 xx 10^(-5))^2) = 2.54 xx 10^(-5) T = 0.254 G` `tan beta = (B_(VR))/(B_(HR)) =(2.24 xx 10^(-5))/(1.19 xx 10^(-5)) = 1.8824` `thereforebeta = tan^(-1) (1.8824)=62^@` |
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