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Q.

The magnetic field due to a current-carrying loop of radius 3 cm at a point on its axis at the distance of 4 cm from the centre is 54µT . What will be its value at the centre of the loop ?

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a

150µT

b

250µT

c

75µT

d

125µT

answer is A.

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Detailed Solution

$\begin{array}{l} B = \frac{μ_{0} {ir}^{2}}{2 {(r^{2} + x^{2})}^{\frac{3}{2}}} \\ 54 \times 10^{- 6} = \frac{μ_{0} i \times 9 \times 10^{- 4}}{2 {[9 \times 10^{- 4} + 16 \times 10^{- 4}]}^{\frac{3}{2}}} \end{array}$
Find μ₀i and substitute in B_C.
$B_{C} = \frac{μ_{0} i}{2 r} we get B_{C} = 250 μT$

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