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Physics Question on Electric Current

An element △l=△xi^\triangle l=\triangle x \hat i is placed at the origin and carries a large current I = 10A. The magnetic field on the y-axis at a distance of 0.5 m from the elements x of 1 cm length is :
The magnetic field on the y-axis at a distance of 0.5 m

A

4×10−8,T4 \times 10^{-8} , \text{T}

B

8×10−8,T8 \times 10^{-8} , \text{T}

C

12×10−8,T12 \times 10^{-8} , \text{T}

D

10×10−8,T10 \times 10^{-8} , \text{T}

Answer

4×10−8,T4 \times 10^{-8} , \text{T}

Explanation

Solution

The position vector for the field point is:

r⃗=0.5j^\vec{r} = 0.5\hat{j}

The magnetic field produced by the current element Δl\Delta l is given by the Biot-Savart law:

dB=μ0I(Δl⃗×r⃗)4πr3dB = \frac{\mu_0 I (\Delta \vec{l} \times \vec{r})}{4\pi r^3}

element vector

Here,

Δl⃗=Δxi^=1100i^ m,r⃗=0.5j^ m,r=∣r⃗∣=0.5 m\Delta \vec{l} = \Delta x \hat{i} = \frac{1}{100}\hat{i} \, \text{m}, \quad \vec{r} = 0.5\hat{j} \, \text{m}, \quad r = |\vec{r}| = 0.5 \, \text{m}

Δl⃗×r⃗=∣i^j^k^ 110000 00.50∣=1100×0.5k^=1200k^\Delta \vec{l} \times \vec{r} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\\ \frac{1}{100} & 0 & 0 \\\ 0 & 0.5 & 0 \end{vmatrix} = \frac{1}{100} \times 0.5 \hat{k} = \frac{1}{200} \hat{k}

Substituting values into the Biot-Savart law:

dB=10−7×10×1200(0.5)3 TdB = \frac{10^{-7} \times 10 \times \frac{1}{200}}{(0.5)^3} \, \text{T}

dB=10−7×10×120018=4×10−8 TdB = \frac{10^{-7} \times 10 \times \frac{1}{200}}{\frac{1}{8}} = 4 \times 10^{-8} \, \text{T}

Hence, the magnetic field is:

dB=4×10−8k^ TdB = 4 \times 10^{-8} \hat{k} \, \text{T}