First slide

Magnetic field due to a current carrying circular ring

Question

A current carrying insulated wire is bent in the form of a circular loop and magnetic field at the centre of the loop is found to be $2 μ T .$ Now the some wire is bent into a loop of three turns. Then new magnetic field at its centre will be _______

Difficult

A

$12 μ T$
B

$18 μ T$
C

$6 μ T$
D

$36 μ T$

Solution

$\begin{array}{l} r = \frac{l}{2 π} ∴ B = \frac{μ_{0} i}{2 r} = \frac{μ_{0} π i}{l} \\ N o w, 2 π r^{'} .2 = l \Rightarrow r^{'} = \frac{l}{2 π n} \\ ∴ B^{'} = \frac{n . μ_{0} i}{2 r^{'}} = \frac{n . μ_{0} i}{l / (π n)} = \frac{μ_{0} π i}{l} {(n)}^{2} \\ = B n^{2} = 2 \times 3^{2} μ T = 18 μ T . \end{array}$

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