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Question: The field B at the centre of a circular coil of radius r is \(\pi\) times that due to a long straig...

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

A

(−π2):(π2):(3π4−12)\left( - \frac { \pi } { 2 } \right) : \left( \frac { \pi } { 2 } \right) : \left( \frac { 3 \pi } { 4 } - \frac { 1 } { 2 } \right)

B

(−π2+1):(π2+1):(3π4+12)\left( - \frac { \pi } { 2 } + 1 \right) : \left( \frac { \pi } { 2 } + 1 \right) : \left( \frac { 3 \pi } { 4 } + \frac { 1 } { 2 } \right)

C

−π2:π2:3π4- \frac { \pi } { 2 } : \frac { \pi } { 2 } : \frac { 3 \pi } { 4 }

D

(−π2−1):(π2−12):(3π4+12)\left( - \frac { \pi } { 2 } - 1 \right) : \left( \frac { \pi } { 2 } - \frac { 1 } { 2 } \right) : \left( \frac { 3 \pi } { 4 } + \frac { 1 } { 2 } \right)

Answer

(−π2):(π2):(3π4−12)\left( - \frac { \pi } { 2 } \right) : \left( \frac { \pi } { 2 } \right) : \left( \frac { 3 \pi } { 4 } - \frac { 1 } { 2 } \right)

Explanation

Solution

Case 1 : BA=μ04π⋅ir⊗\mathrm { B } _ { \mathrm { A } } = \frac { \mu _ { 0 } } { 4 \pi } \cdot \frac { \mathrm { i } } { \mathrm { r } } \otimes

BB=μ04π⋅irB _ { B } = \frac { \mu _ { 0 } } { 4 \pi } \cdot \frac { i } { r }◉

BC=μ04π⋅irB _ { C } = \frac { \mu _ { 0 } } { 4 \pi } \cdot \frac { i } { r }◉

So net magnetic field at the centre of case 1

B1=BB−(BA+BC)B _ { 1 } = B _ { B } - \left( B _ { A } + B _ { C } \right)

⇒ B1=μ04π⋅πirB _ { 1 } = \frac { \mu _ { 0 } } { 4 \pi } \cdot \frac { \pi i } { r }◉ ..... (i)

Case 2 :

As we discussed before magnetic field at the centre O in this case

B2=μ04π⋅πir⊗B _ { 2 } = \frac { \mu _ { 0 } } { 4 \pi } \cdot \frac { \pi i } { r } \otimes .....(ii)

Case 3 : BA=0B _ { A } = 0

BB=μ04π⋅(2π−π/2)r⊗B _ { B } = \frac { \mu _ { 0 } } { 4 \pi } \cdot \frac { ( 2 \pi - \pi / 2 ) } { r } \otimes =μ04π⋅3πi2r⊗= \frac { \mu _ { 0 } } { 4 \pi } \cdot \frac { 3 \pi i } { 2 r } \otimes

BC=μ04π⋅irB _ { C } = \frac { \mu _ { 0 } } { 4 \pi } \cdot \frac { i } { r }◉

So net magnetic field at the centre of case 3

B3=μ04π⋅ir(3π2−1)⊗B _ { 3 } = \frac { \mu _ { 0 } } { 4 \pi } \cdot \frac { i } { r } \left( \frac { 3 \pi } { 2 } - 1 \right) \otimes ....(iii)

From equation (i), (ii) and (iii) =−π2:π2:(3π4−12)= - \frac { \pi } { 2 } : \frac { \pi } { 2 } : \left( \frac { 3 \pi } { 4 } - \frac { 1 } { 2 } \right)