The magnetic field 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. Figure shows three cases. In all cases the circular path has radius r and straight ones are infinitely long. For same current the B field at the centre P in cases 1,2,3 has the ratio :

- (\frac{π}{4} + \frac{1}{2}) : (\frac{π}{2}) : (\frac{3 π}{4} - \frac{1}{2})

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(- \frac{π}{2} + 1) : (\frac{π}{2} + 1) : (\frac{3 π}{4} + \frac{1}{2})

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- \frac{π}{2} : \frac{π}{2} : \frac{3 π}{4}

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(- \frac{π}{2} - 1) : (\frac{π}{2} - \frac{1}{4}) : (\frac{3 π}{4} + \frac{1}{2})

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Solution

The correct option is B $- (\frac{π}{4} + \frac{1}{2}) : (\frac{π}{2}) : (\frac{3 π}{4} - \frac{1}{2})$
Magnetic field in case 1 is $B_{1} = \frac{μ_{0} I}{2 r} (\frac{π}{2 \times 2 π} + \frac{1}{2 π})$ outside the plane of paper
Magnetic field in case 2 is $B_{2} = \frac{μ_{0} I}{4 π r}$ inside the plane of paper
Magnetic field in case 3 is $B_{3} = \frac{μ_{0} I}{2 r} (\frac{3 π}{2 \times 2 π} - \frac{1}{2 π})$ inside the plane of paper
$∴ B_{1} : B_{2} : B_{3}$ = $- (\frac{π}{4} + \frac{1}{2}) : (\frac{π}{2}) : (\frac{3 π}{4} - \frac{1}{2})$

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Similar questions

(1 + \frac{1}{2}) (1 + \frac{1}{3}) (1 + \frac{1}{4}) . . . . . . . . . . . . . . (1 + \frac{1}{n}) =

(\frac{4}{49})^{\frac{3}{4}} \times (\frac{49}{4})^{- \frac{1}{2}} .

f (x)

= {\begin{matrix} 2 x + 1; x \leq 0 x^{2}; x > 0 \end{matrix}

f (\frac{1}{2}), f (\frac{- 1}{2})

f (x)

= {\begin{matrix} {sin}^{2} x; x \leq 0 {cos}^{2} x; x > 0 \end{matrix}

f (\frac{π}{3}) + f (\frac{- π}{2})

^{'} a^{'}

^{'} b^{'}

f (x)

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} x + a \sqrt{2} sin x, & 0 \leq x < \frac{π}{4} 2 x cot x + b, & \frac{π}{4} \leq x \leq \frac{π}{2} a cos 2 x - b sin x, & \frac{π}{2} < x \leq π \end{matrix}

x = \frac{π}{4}, \frac{π}{2}

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