A current carrying loop ABCD has two circular arcs AD and BC with radius 1 cm and 2 cm respectively as shown in the figure. The two arcs AD and BC subtend a common angle $30^{0}$ at the centre O. If the current flowing in the loop is $\frac{1.2}{π} A$ , then the magnitude of net magnetic field at O is $(Given μ_{0} = 4 π \times 10^{- 7})$

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$3 μ T$

$1 μ T$

$1.5 μ T$

$0.5 μ T$

answer is C.

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Detailed Solution

$B_{0} {= B}_{AB} {+ B}_{BC} {+B}_{CD} {+B}_{DA}$

$= 0 + \frac{μ_{0} i}{24 × 0. 02} \otimes + 0 + \frac{μ_{0} i}{24 × 0. 01} ⊙$

$= \frac{μ_{0} i}{24} [\frac{1}{0. 02} - \frac{1}{0. 01}] \otimes$

$= \frac{{4π × 10}^{-7}}{24} × \frac{1 . 2}{π} [50-100] \otimes$

$= \frac{4 \times 10^{- 7} \times 1. 2}{24} \times 50 \otimes = - 1 μ T \otimes = 1 μ T ⊙$

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