Question

The magnetic moment of a short dipole is, $1{\text{Am}}^2$. What is the magnitude of the magnetic induction in air at $10\text{cm}$ from the center of the dipole on a line making an angle of ${30}^{\circ }$ with the axis of the dipole?

• A. $\frac{\sqrt{13}}{2}\times {10}^{-4}\text{T}$
• B. $\frac{\sqrt{13}}{4}\times {10}^{-4}\text{T}$
• C. $\sqrt{\frac{13}{2}}\times {10}^{-4}\text{T}$
• D. $\sqrt{13}\times {10}^{-4}\text{T}$

Answer 1

The correct option is A $\frac{\sqrt{13}}{2}\times {10}^{-4}\text{T}$

Given, $M=1{\text{Am}}^2;d=10\text{cm};\theta ={30}^{\circ }$

Magnetic induction at any arbitrary point due to a magnetic dipole moment is,

$B=\frac{{\mu }_{0}M}{4\pi d^3}\sqrt{1+3{\cos }^2\theta }$

$B=\frac{4\pi \times {10}^{-7}\times 1}{4\pi \times {10}^{-3}}\sqrt{1+3{\left(\frac{\sqrt{3}}{2}\right)}^2}$

$B=\frac{\sqrt{13}}{2}\times {10}^{-4}\text{T}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.