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A current I enters a uniform circular loop of radius R at point M and flows out at N as shown in the Fig. Obtain the net magnetic field at the centre O of the loop. |
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Answer» Solution :The net magnet field at centreO of the loop is vector sum of fields due to TWO current segments 1 and 2 carrying currents `I_1 and I_2` and having resistance `R_1 and R_2` respectively. If V be the potential difference between M and N , then `V = I_1 R_1 = I_2 R_2` If `R_0` be the total resistance of entire loop than `R_1 = (90^@)/(360^(@)) R_0 = (R_0)/(4) and R_2 = R_0 - R_1` and `R_0 - (R_0)/(4) = (3 R_0)/(4)` `implies I_2 = (I_1 R_1)/(R_2) = (l_1 (R_0//4))/((3 R_0//4)) = (l_1)/3` `:.` Field `vec(B_1)` due to current SEGMENT (1), `vec(B_1) = (mu_0 I_1)/(2 R) cdot 1/4 OX` and field `vec(B_2)` due to current segment (2), `vec(B_2) = (mu_0 I_2)/(2 R)cdot 3/4 o. = (mu_0)/(2R)cdot (l_i)/(3)cdot 3/4 o.` `implies vec(B_2) = (mu_0 I_1)/(2R) cdot 1/4 o.` `:.` Net magnetic field at 0, `vecB = vec(B_1) + vec(B_2) = (mu_0 I_1)/(8 R) ox + (mu_0 I_1)/(8R) o. = vec0` 
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