Question

Two short bar magnets of moments 1.08Am${}^2$ and 1.92 Am${}^2$ are placed along two lines drawn on the table at right angles to each other and the centres of the magnets are respectively at 0.3m and 0.4m from the point of intersection. The magnetic induction at the point of intersection of the lines is :

• A. ${10}^{-4}$ T
• B. ${10}^{-5}$ T
• C. $14\times {10}^{-5}T$
• D. $0.35\times {10}^{-6}T$

Answer 1

The correct option is B ${10}^{-5}$ T

$m_1=1.08A{m}^2$
$m_2=1.92A{m}^2$
$B_1=\frac{2{\mu }_0{m}_1}{4\pi d_1^3}$
$B_2=\frac{2{\mu }_0{m}_2}{4\pi d_2^3}$
$B=\sqrt{B_1^2+B_2^2}$
$=\sqrt{{\left(\frac{2\times {10}^{-7}\times 1.08}{{0.3}^3}\right)}^2+{\left(\frac{2\times {10}^{-7}\times 1.92}{{0.4}^3}\right)}^2}$
$=\sqrt{{(80\times {10}^{-7})}^2+{(60\times {10}^{-7})}^2}$
$=100\times {10}^{-7}$
$={10}^{-5}T$