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Physics-

General

Easy

Question

The unit vectors $stack i with hat on top comma stack blank j with hat on top$ and $stack k with hat on top$ are as shown below. What will be the magnetic field at $O$ in the following figure

$\frac{μ_{0}}{4 π} \frac{i}{a} (2 - \frac{π}{2}) \hat{j}$
$\frac{μ_{0}}{4 π} \frac{i}{a} (2 + \frac{π}{2}) \hat{j}$
$\frac{μ_{0}}{4 π} \frac{i}{a} (2 + \frac{π}{2}) \hat{i}$
$\frac{μ_{0}}{4 π} \frac{i}{a} (2 + \frac{π}{2}) \hat{k}$

The correct answer is: $fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction fraction numerator i over denominator a end fraction open parentheses 2 plus fraction numerator pi over denominator 2 end fraction close parentheses stack k with hat on top$

The field at $O$ due to $A B$ is $fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction. fraction numerator i over denominator a end fraction stack k with hat on top$ and that due to $D E$ is also $fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction. fraction numerator i over denominator a end fraction stack k with hat on top$
However the field due t $B C D$ is $fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction. fraction numerator i over denominator a end fraction open parentheses fraction numerator pi over denominator 2 end fraction close parentheses stack k with hat on top$
Thus the total field at $O$ is $fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction. fraction numerator i over denominator a end fraction open parentheses 2 plus fraction numerator pi over denominator 2 end fraction close parentheses stack k with hat on top$

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