The unit vectors hat{i}, hat{j}, and hat{k} a | vedclass

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Question

(D) The wire consists of three parts: a semi-infinite straight wire $AB$,a quarter-circular arc $BCD$,and another semi-infinite straight wire $DE$.$1$

Vedclass · Accepted Answer

$\frac{\mu_0}{4\pi} \frac{i}{a} \left( 2 + \frac{\pi}{2} \right) \hat{k}$

Vedclass · Answer

(D) The wire consists of three parts: a semi-infinite straight wire $AB$,a quarter-circular arc $BCD$,and another semi-infinite straight wire $DE$.$1$. For the semi-infinite wire $AB$,the magnetic field at $O$ is given by $B_1 = \frac{\mu_0 i}{4\pi a}$. Using the right-hand rule,the direction is $\hat{k}$.$2$. For the semi-infinite wire $DE$,the magnetic field at $O$ is given by $B_2 = \frac{\mu_0 i}{4\pi a}$. Using the right-hand rule,the direction is $\hat{k}$.$3$. For the quarter-circular arc $BCD$ of radius $a$,the magnetic field at the center is $B_3 = \frac{\mu_0 i}{4a} 	imes \frac{1}{4} = \frac{\mu_0 i}{4\pi a} 	imes \frac{\pi}{2}$. The direction is $\hat{k}$.$4$. The total magnetic field at $O$ is the vector sum: $B_{total} = B_1 + B_2 + B_3 = \frac{\mu_0 i}{4\pi a} \left( 1 + 1 + \frac{\pi}{2} ight) \hat{k} = \frac{\mu_0}{4\pi} \frac{i}{a} \left( 2 + \frac{\pi}{2} ight) \hat{k}$.

The unit vectors $\hat{i}, \hat{j},$ and $\hat{k}$ are as shown below. What will be the magnetic field at $O$ in the following figure?

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