Find the magnetic field at point P due to a straight line segment AB of length 6 cm carrying a current of 5 A. (see figure) ($\mu _{0}=4\pi \times 10^{-7}N-A^{-2})$ Option 1) $2.0\times 10^{-5}T$       Option 2) $1.5\times 10^{-5}T$ Option 3) $3.0\times 10^{-5}T$ Option 4) $2.5\times 10^{-5}T$

Magnetic Field Due to a Straight Wire -

$B=\frac{\mu_{o}i}{4\pi r}(\sin\phi_{1}+\sin\phi_{2})$

- wherein

$\sin \Theta =\frac{3}{5}$

So $B_{p}=\frac{\mu _{0}I}{4\pi d}\times 2\sin \Theta$

$B_{p}=\frac{10^{7}\times 4\pi }{4\pi }\times \frac{5\times 2}{4\times 10^{-2}}\times \frac{3}{5}$

$B_{p}=1.5\times 10^{-5}T$

Option 1)

$2.0\times 10^{-5}T$

Option 2)

$1.5\times 10^{-5}T$

Option 3)

$3.0\times 10^{-5}T$

Option 4)

$2.5\times 10^{-5}T$

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