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) ($$\mu_0 = 4\pi \times 10^{-7}$$ NA$$^{-2}$$)

Perpendicular distance ($$d$$) from wire to point P: $$d = \sqrt{5^2 - 3^2} = 4\text{ cm} = 0.04\text{ m}$$

Angles from perpendicular to wire ends: $$\sin\theta_1 = \sin\theta_2 = \frac{3}{5}$$

Magnetic field calculation: $$B = \frac{\mu_0 I}{4\pi d}(\sin\theta_1 + \sin\theta_2)$$

$$B = \frac{10^{-7} \times 5}{0.04} \left(\frac{3}{5} + \frac{3}{5}\right) = \frac{5 \times 10^{-7}}{0.04} \times \frac{6}{5} = \frac{6 \times 10^{-7}}{0.04} = 1.5 \times 10^{-5}\text{ T}$$