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Q.

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

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a

−π2:π2:3π4−12

b

−π2+1:π2+1:3π4+12

c

−π2:π2:3π4

d

−π2−1:π2−14:3π4+12

answer is A.

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Detailed Solution

(1) B1=μ0I4πr=B3⊗⊙B2=μ0I4rB→net =B→1+B→2+B→3Bnet =μ0I4r⊙ … (i) B1=B3=0 B→net =B→1+B→2+B→3⊗B2=μ0I4R=μ0I4r⊗ … (ii) (3) B1=0⊗B2=μ0I4πr3π2=3μ0I8r⊙B3=μ0I4πrB→net =μ0I4r32−1π⊗ … (iii) B(1):B(2):B(3)−1:1:32−1π⇒ −π2:π2:3π4−12

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