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An infinitely long wire carrying a current i is bent at its mid point 0 to form an angle45^@. P is point a distance Rfrom the pointofbending. Find the magnetic field at P. |
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Answer» <P> Solution :Since point P lies on the axis of straight part OA, magnetic field at point P is zero due to path OA of wire From `Delta OPN, d= R cos 45^0` Since both the ends O and C are on the same side of normal PN, `phi_1 = -45^0 and phi_2 = + 90^0` So `B = (mu_0 i)/(4pid) (sin phi_1 + sin phi_2)` ` = (mu_0i)/(4PI R cos 45^0)[sin (-45^0) + sin 90^0]` ` = (mu_0 i XX sqrt(2))/(4piR) (-sin45^@ + 1), = (sqrt(mu_0i))/(4piR)(1- (1)/(sqrt(2)))OX` `= ((sqrt(2)-1)mu_0i)/(4piR) ox` (into the page) |
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