Q 12-04-228JEE MainJEE Main 2019 (12 Apr, Shift 2)Medium
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). ($\mu_0 = 4\pi\times10^{-7}\ \text{N A}^{-2}$)
Answer: (A) $1.5\times10^{-5}$ T
Perpendicular distance from P to AB: $d = \sqrt{5^2 - 3^2} = 4$ cm. Each end subtends an angle $\alpha$ with $\sin\alpha = \dfrac35$.
$$B = \frac{\mu_0I}{4\pi d}(\sin\alpha_1 + \sin\alpha_2) = \frac{10^{-7}\times5}{0.04}\times\frac65 = 1.5\times10^{-5}\ \text{T}$$
Solution by Sreeraj P, M.Sc Physics