A coil of wire 20 cm length and 200 turns carries a current of 10 A.A bar of titanium (magnetic susceptibility 1.81 x 10^-4)is positioned within the coil.The magnetic field inside the titanium bar is A)7.590 x 10^-4 T. B)4.145 x 10^-4 T. C)12.6 x 10^-4 T. D)1.257 x 10^-2 T.

Question

Quizplus · Accepted Answer

The magnetic field inside the titanium bar can be calculated using the formula for the magnetic field inside a material, which is $B = \mu H$, where $\mu = \mu_0(1 + \chi)$ and $H$ is the magnetic field due to the current in the coil. For a coil, $H = nI$, where $n$ is the number of turns per unit length and $I$ is the current. Given that the coil has 200 turns and a length of 20 cm (0.2 m), $n = 200 / 0.2 = 1000$ turns/m. With $I = 10$ A, $H = 1000 \times 10 = 10000$ A/m. The magnetic susceptibility of titanium ($\chi$) is $1.81 \times 10^{-4}$, and $\mu_0 = 4\pi \times 10^{-7}$ Tm/A. Thus, $\mu = 4\pi \times 10^{-7} \times (1 + 1.81 \times 10^{-4})$, and the magnetic field $B = \mu H = 4\pi \times 10^{-7} \times (1 + 1.81 \times 10^{-4}) \times 10000$, which calculates to approximately $1.257 \times 10^{-2}$ T, matching option D.

A Coil of Wire 20 Cm Length and 200 Turns