In the figure,find out the magnetic field at | vedclass

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(A) The magnetic field at the center $B$ is the sum of the magnetic fields due to the straight wire segments and the circular loop.For a straight wire

Vedclass · Accepted Answer

$\pi 	imes \left[1 + \frac{1}{\pi}ight] 	imes 10^{-5} \; T$

Vedclass · Answer

(A) The magnetic field at the center $B$ is the sum of the magnetic fields due to the straight wire segments and the circular loop.For a straight wire segment of infinite length,the field at distance $r$ is $B = \frac{\mu_0 I}{2 \pi r}$.For a circular loop of radius $r$,the field at the center is $B = \frac{\mu_0 I}{2 r}$.In the given configuration,the magnetic field at the center due to the straight parts and the loop all point in the same direction (into the page).The total magnetic field is $B_{total} = B_{wire} + B_{loop} = \frac{\mu_0 I}{2 \pi r} + \frac{\mu_0 I}{2 r} = \frac{\mu_0 I}{2 r} \left[ \frac{1}{\pi} + 1 ight]$.Substituting the values: $\mu_0 = 4 \pi 	imes 10^{-7} \; T \cdot m/A$,$I = 2.5 \; A$,$r = 5 	imes 10^{-2} \; m$.$B_{total} = \frac{(4 \pi 	imes 10^{-7}) 	imes 2.5}{2 	imes 5 	imes 10^{-2}} \left[ \frac{1}{\pi} + 1 ight] = \frac{10 \pi 	imes 10^{-7}}{10 	imes 10^{-2}} \left[ \frac{1}{\pi} + 1 ight] = \pi 	imes 10^{-5} \left[ \frac{1}{\pi} + 1 ight] \; T$.

In the figure,find out the magnetic field at the center $B$ (Given $I = 2.5 \; A, r = 5 \; cm$).

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