A long straight wire carries a current of 35; | vedclass

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(C) Current in the wire,$I = 35 \; A$.Distance of a point from the wire,$r = 20 \; cm = 0.2 \; m$.The magnitude of the magnetic field $B$ at a distanc

Vedclass · Accepted Answer

$3.5 	imes 10^{-5} \;T$

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(C) Current in the wire,$I = 35 \; A$.Distance of a point from the wire,$r = 20 \; cm = 0.2 \; m$.The magnitude of the magnetic field $B$ at a distance $r$ from a long straight wire is given by the formula:$B = \frac{\mu_0}{4\pi} \cdot \frac{2I}{r}$Where $\frac{\mu_0}{4\pi} = 10^{-7} \; T \cdot m/A$.Substituting the values:$B = 10^{-7} 	imes \frac{2 	imes 35}{0.2}$$B = 10^{-7} 	imes \frac{70}{0.2}$$B = 10^{-7} 	imes 350$$B = 3.5 	imes 10^{-5} \; T$.Thus,the magnitude of the magnetic field is $3.5 	imes 10^{-5} \; T$.

$A$ long straight wire carries a current of $35\; A$. What is the magnitude of the field $B$ at a point $20\; cm$ from the wire?

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