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A compass needle free to turn in a horizontal plane is placed at the centre of circular coil of 30 turns and radius 12 cm. The coil is in a vertical plane making an angle of 45^@ with the magnetic meridian. When the current in the coil is 0.35 A, the needle points west to east. (a) Determine the horizontal component of the earth's magnetic field at the location. (b) The current in the coil is reversed, and the coil is rotated about its vertical axis by an angle of 90^@ in the anticlockwise sense looking from above. Predict the direction of the needle. Take the magnetic declination at the places to be zero. |
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Answer» Solution :Here N= 30 , R= 12 cm = 0.12 m , I = 0.35 A Let magnetic field due to current FLOWING in the coil be B given by, `B= (mu_0 NI)/(2R)` and is DIRECTED normal to be plane of coil as shown in Fig. The field B can be resolved into two components , namely B cos `45^@` along W-E DIRECTION into two componenets , namely B cos `45^@` along W-E direction and B sin `45^@` along N-S direction. The magnetic needle can point west to east only when component B sin `45^@` is just balanced by `B_H`, the horizontal component of earth.s magnetic field at that place . Thus, `B_H = B sin 45^@ = (mu NI)/(2R) sin 45^@` `therefore B_H = (4pi xx 10^(-7) xx 30xx0.35)/(2xx 0.12) xx 1/(sqrt2) = 3.9 xx 10^(-5) T. ` (b) When current in the coil is reversed and the coil is turned through `90^@`in the anticlockwise sense looking from above, the direction of magnetic needle will get reversed i.e. it will point from east to west . 
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