Q152 / 160 · AP EAMCET 2018 (22 Apr Shift 1) · Physics
Two short bar magnets each of magnetic moment of $9 \mathrm{Am}^2$ are placed such that one is at $x=-3 \mathrm{~cm}$ and the other at $y=-3 \mathrm{~cm}$. If their magnetic moments are directed along positive and negative $X$-directions respectively, then the resultant magnetic field at the origin is
- A100T
- B10T
- Correct answerC0.1T
- D0.001T
Solution
At origin, both magnetic fields will be in same direction.
Magnetic field due to $M_1$ (axial point) $$ \begin{aligned} & B_1=\frac{\mu_0}{4 \pi} \times \frac{2 M}{r^3}=10^{-7} \times \frac{2 \times 9}{27 \times 10^{-6}} \\ & B_1=\frac{2}{3} \times 10^{-1} \mathrm{~T} \end{aligned} $$ Magnetic field due to $M_2$ (equatorial point) $$ B_2=\frac{\mu_0}{4 \pi} \times \frac{M}{r^3}=10^{-7} \times \frac{9}{27 \times 10^{-6}}=\frac{1}{3} \times 10^{-1} \mathrm{~T} $$ As both $B_1$ and $B_2$ points in same direction, so effective field, $B=\left(\frac{1}{3}+\frac{2}{3}\right) \times 10^{-1} \mathrm{~T}$ $$ \Rightarrow \quad B=10^{-1} \mathrm{~T}=0.1 \mathrm{~T} $$
Magnetic field due to $M_1$ (axial point) $$ \begin{aligned} & B_1=\frac{\mu_0}{4 \pi} \times \frac{2 M}{r^3}=10^{-7} \times \frac{2 \times 9}{27 \times 10^{-6}} \\ & B_1=\frac{2}{3} \times 10^{-1} \mathrm{~T} \end{aligned} $$ Magnetic field due to $M_2$ (equatorial point) $$ B_2=\frac{\mu_0}{4 \pi} \times \frac{M}{r^3}=10^{-7} \times \frac{9}{27 \times 10^{-6}}=\frac{1}{3} \times 10^{-1} \mathrm{~T} $$ As both $B_1$ and $B_2$ points in same direction, so effective field, $B=\left(\frac{1}{3}+\frac{2}{3}\right) \times 10^{-1} \mathrm{~T}$ $$ \Rightarrow \quad B=10^{-1} \mathrm{~T}=0.1 \mathrm{~T} $$
Hence, the correct answer is option: 'C'.