Magnetic Effects Of Current Question 133
The magnetic field at the centre of a circular coil of radius r is $ \frac{\mu_0 I}{2r} $ 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]
Options:
A) $ ( -\frac{\pi }{2} ):( \frac{\pi }{2} ):( \frac{3\pi }{4}-\frac{\pi }{2} ) $
B) $ ( -\frac{\pi }{2}+1 ):( \frac{\pi }{2}+1 ):( \frac{3\pi }{4}+\frac{1}{2} ) $
C) $ -\frac{\pi }{2}:\frac{\pi }{2}:\frac{3\pi }{4} $
D) $ ( -\frac{\pi }{2}-1 ):( \frac{\pi }{2}-\frac{1}{4} ):( \frac{3\pi }{4}+\frac{1}{2} ) $
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Answer:
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}( \frac{3\pi }{2}-1 )\otimes $ ….. (iii) From equation (i), (ii) and (iii) $ B_{1}:B_{2}:B_{3}=\pi $ ¤ : $ \pi \otimes $ $ ( \frac{3\pi }{2}-1 ),\otimes =-\frac{\pi }{2}:\frac{\pi }{2}:( \frac{3\pi }{4}-\frac{1}{2} ) $
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