First slide

Magnetic field due to a straight current carrying wire

Question

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

Moderate

A

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

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

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

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

Solution

In fig.(a) the magnetic fie]d induction at point P is given by
$B_{1} = - \frac{B}{2} - \frac{πB}{2} + \frac{B}{2} = - \frac{πB}{2}$ …………………(1)
In fig. (b) the ma8netic induction at point P is given by
$B_{2} = \frac{πB}{2}$ ……………………………………………(2)
In fig. (c), the magnetic field induction at point P is given by
$B_{3} = \frac{3 πB}{4} - \frac{B}{2} = (\frac{3 π}{4} - \frac{1}{2}) B$ ̣………………….(3)
$\begin{aligned} ∴ B_{1} : B_{2} : B_{3} = - \frac{πB}{2} : \frac{πB}{2} : (\frac{3 π}{4} - \frac{1}{2}) B \\ = - \frac{π}{2} : \frac{π}{2} : (\frac{3 π}{4} - \frac{1}{2}) \end{aligned}$

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