TG EAMCET 4 May 2025 Shift 1 Paper

Step 1: Find the magnetic field due to the first wire $(8 \text{ A})$The formula for the magnetic field at a distance $r$ from a straight wire carrying current $I$ is:$B = \frac{\mu_0 I}{2 \pi r}$For the first wire: $I_1 = 8 \text{ A}, r_1 = 4 \text{ cm} = 4 \times 10^{-2} \text{ m}$So,$B_1 = \frac{\mu_0 \times 8}{2 \pi \times 4 \times 10^{-2}} (-\hat{k}) \text{ (The } -\hat{k} \text{ shows the direction of the field.) }$Step 2: Find the magnetic field due to the second wire (10 A)The second wire has $I_2 = 10 \text{ A}$ and is $r_2 = 5 \text{ cm} = 5 \times 10^{-2} \text{ m}$ away from the point. (Since the wires are 9 cm apart and the point is 4 cm from the first wire, it must be 5 cm from the second wire.)So, $B_{\text{II}} = \frac{\mu_0 \times 10}{2 \pi \times 5 \times 10^{-2}} (-\hat{k})$Step 3: Add the two magnetic fieldsBoth fields point in the same direction $(-\hat{k})$, so add their values:$B_{\text{net}} = B_1 + B_{\text{II}} = \frac{\mu_0}{2 \pi \times 10^{-2}} \left[ \frac{8}{4} + \frac{10}{5} \right] (-\hat{k})$Calculate inside the brackets:$\frac{8}{4}=2, \frac{10}{5}=2$So, $B_{\text{net}} = \frac{\mu_0}{2 \pi \times 10^{-2}} [2+2] (-\hat{k}) = \frac{\mu_0}{2 \pi \times 10^{-2}} \cdot 4 (-\hat{k})$Step 4: Substitute the value of $\mu_0$We know $\mu_0 = 4 \pi \times 10^{-7} \text{ Tm/A}$.So, $B_{\text{net}} = \frac{4 \pi \times 10^{-7}}{2 \pi \times 10^{-2}} \cdot 4 (-\hat{k})$The $\pi$ cancels: $\frac{4}{2}=2$$\text{ So, } = \frac{2 \times 10^{-7} \times 4}{10^{-2}} (-\hat{k}) = \frac{8 \times 10^{-7}}{10^{-2}} (-\hat{k}) = 8 \times 10^{-5} \text{ T} (-\hat{k})$