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Directions : Question are based on the following paragraph.

A current loop $ABCD$ is held fixed on the plane of the paper as shown in the figure. The arcs $BC(radius=b)$ and $DA(radius=a)$ of the loop are joined by two straight wires $AB$ and $CD$ . A steady current $I$ is flowing in the loop. Angle made by $AB$ and $CD$ at the origin $O$ is $30^{\circ}$ . Another straight thin wire with steady current $I_{1}$ flowing out of the plane of the paper is kept at the origin.

Question: The magnitude of the magnetic field $(B)$ due to loop $ABCD$ at the origin $(O)$ is

Option 1)
zero

Option 2)
$\frac{\mu_{0} I(b-a)}{24ab}\;\; \;$

Option 3)
$\; \frac{\mu _{0}I}{4\pi }\left [ \frac{b-a}{ab} \right ]\;$

Option 4)
$\; \; \frac{\mu _{0}I}{4\pi }\left [ 2(b-a)+\frac{\pi }{3}(a+b) \right ]$

Answers (1)

best_answer

As we learnt in

Arc subtends angle theta at the centre -

$B=\frac{\mu_{0}}{4\pi }.\frac{\theta i}{r}$

- wherein

$B_1 =\frac{\mu_0I}{4\pi a}\times \frac{\pi }{6} (directed\:vertically\:upward)$

$B_2 =\frac{\mu_0I}{4\pi b}\times \frac{\pi }{6} (directed\:vertically\:downward)$

$B=B_1 -B_2 =\frac{\mu_0I}{24}\left [ \frac{1}{a}-\frac{1}{b} \right ]=\frac{\mu_0I}{24ab} (b-a)$

Option 1)

zero

Incorrect

Option 2)

$\frac{\mu_{0} I(b-a)}{24ab}\;\; \;$

Correct

Option 3)

$\; \frac{\mu _{0}I}{4\pi }\left [ \frac{b-a}{ab} \right ]\;$

Incorrect

Option 4)

$\; \; \frac{\mu _{0}I}{4\pi }\left [ 2(b-a)+\frac{\pi }{3}(a+b) \right ]$

Incorrect

Posted by

prateek

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