The wire shown in figure carries a current of $40 A$ . If $r = 3.14 c m$ the magnetic field at point p will be :

Question

The wire shown in figure carries a current of 40A- If r-3-14cm the magnetic field point p will be -1-6times 10-3-T3-2times 10-3-T4-8times 10-3-T6times 10-4-T

Toppr Admin · Accepted Answer

Magnetic field due to circular arc is given by-B-x3BC-0I4-x3C0-r-x3B8- -x3BC-0I-xD7-3-x3C0-4-x3C0-r-xD7-2-3-x3BC-0I8r-3-xD7-4-x3C0-xD7-10-x2212-7-xD7-408-xD7-3-14-xD7-10-x2212-2-6-xD7-10-x2212-4T

The wire shown in figure carries a current of $40 A$ . If $r = 3.14 c m$ the magnetic field at point p will be :

$1.6 \times 10^{- 3} T$
$3.2 \times 10^{- 3} T$
$4.8 \times 10^{- 3} T$
$6 \times 10^{- 4} T$

Magnetic field due to circular arc is given by,
$| B | = \frac{μ_{0} I}{4 π r} θ$ = $\frac{μ_{0} I \times 3 π}{4 π r \times 2}$ = $\frac{3 μ_{0} I}{8 r}$
= $\frac{3 \times 4 π \times 10^{- 7} \times 40}{8 \times 3.14 \times 10^{- 2}}$
= $6 \times 10^{- 4} T$

The wire shown in figure carries a current of 40A. If r=3.14cm the magnetic field at point p will be :1.6×10−3T3.2×10−3T4.8×10−3T6×10−4T

Magnetic field due to circular arc is given by,|B|=μ0I4πrθ =μ0I×3π4πr×2=3μ0I8r=3×4π×10−7×408×3.14×10−2=6×10−4T

The wire shown in figure carries a current of $40 A$ . If $r = 3.14 c m$ the magnetic field at point p will be :

$1.6 \times 10^{- 3} T$
$3.2 \times 10^{- 3} T$
$4.8 \times 10^{- 3} T$
$6 \times 10^{- 4} T$

Magnetic field due to circular arc is given by,
$| B | = \frac{μ_{0} I}{4 π r} θ$ = $\frac{μ_{0} I \times 3 π}{4 π r \times 2}$ = $\frac{3 μ_{0} I}{8 r}$
= $\frac{3 \times 4 π \times 10^{- 7} \times 40}{8 \times 3.14 \times 10^{- 2}}$
= $6 \times 10^{- 4} T$