Question 13

An element $$\Delta l = \Delta x\hat{i}$$ is placed at the origin and carries a large current $$I = 10 \text{ A}$$. The magnetic field on the $$y$$-axis at a distance of $$0.5 \text{ m}$$ from the element of length $$\Delta x$$ of $$1 \text{ cm}$$ is:

$$\vec{dl} = \Delta x \hat{i} = 10^{-2}\hat{i}\ \text{m}$$

$$\vec{r} = 0.5\hat{j}\ \text{m} \implies \hat{r} = \hat{j}$$

$$\vec{dB} = \frac{\mu_0}{4\pi} \frac{I (\Delta x \hat{i} \times \hat{j})}{r^2}$$

$$dB = 10^{-7} \times \frac{10 \times 10^{-2} \times \sin(90^\circ)}{(0.5)^2}$$

$$dB = 10^{-7} \times \frac{10^{-1}}{0.25} = 4 \times 10^{-8}\ \text{T}$$

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