question_answer

The magnetic field at the centre of a circular coil of radius r is \[\pi \] times that due to a long straight wire at a distance r from it, for equal currents. Figure here shows three cases : in all cases the circular part has radius r and straight ones are infinitely long. For same current the B field at the centre P in cases 1, 2, 3 have the ratio                                         [CPMT 1989]

A)            \[\left( -\frac{\pi }{2} \right):\left( \frac{\pi }{2} \right):\left( \frac{3\pi }{4}-\frac{1}{2} \right)\]

B)            \[\left( -\frac{\pi }{2}+1 \right):\left( \frac{\pi }{2}+1 \right):\left( \frac{3\pi }{4}+\frac{1}{2} \right)\]

C)            \[-\frac{\pi }{2}:\frac{\pi }{2}:3\frac{\pi }{4}\]

D)            \[\left( -\frac{\pi }{2}-1 \right):\left( \frac{\pi }{2}-\frac{1}{4} \right):\left( \frac{3\pi }{4}+\frac{1}{2} \right)\]

Correct Answer: A

Solution :

                    Case 1: \[{{B}_{A}}=\frac{{{\mu }_{0}}}{4\pi }.\frac{i}{r}\otimes \] \[{{B}_{B}}=\frac{{{\mu }_{0}}}{4\pi }.\frac{\pi \,i}{r}\]¤ \[{{B}_{C}}=\frac{{{\mu }_{0}}}{4\pi }.\frac{i}{r}\]¤ So net magnetic field at the centre of case 1 \[{{B}_{1}}={{B}_{B}}-{{B}_{C}}-{{B}_{A}}\Rightarrow {{B}_{1}}=\frac{{{\mu }_{0}}}{4\pi }.\frac{\pi i}{r}\]¤   ..... (i) Case 2 : As we discussed before magnetic field at the centre O in this case \[{{B}_{2}}=\frac{{{\mu }_{0}}}{4\pi }.\frac{\pi i}{r}\otimes \] ..... (ii) Case 3 : \[{{B}_{A}}=0\] \[{{B}_{B}}=\frac{{{\mu }_{0}}}{4\pi }.\frac{(2\pi -\pi /2)i}{r}\otimes \] \[{{B}_{C}}=\frac{{{\mu }_{0}}}{4\pi }.\frac{i}{r}\]¤ \[=\frac{{{\mu }_{0}}}{4\pi }.\frac{3\pi i}{2r}\otimes \] So net magnetic field at the centre of case 3 \[{{B}_{3}}=\frac{{{\mu }_{0}}}{4\pi }.\frac{i}{r}\left( \frac{3\pi }{2}-1 \right)\otimes \] ..... (iii) From equation (i), (ii) and (iii) \[{{B}_{1}}:{{B}_{2}}:{{B}_{3}}=\pi \]¤ : \[\pi \otimes \] \[\left( \frac{3\pi }{2}-1 \right)\,\otimes =-\frac{\pi }{2}:\frac{\pi }{2}:\left( \frac{3\pi }{4}-\frac{1}{2} \right)\]