Question

Two magnetic poles have their strengths in the ratio 3 : 2. They are kept at a distance of 0.6 m in air and the force of repulsion is found to be 0.06 dynes. The pole strengths are (in amp. m)

• A. 1.8, 1.2
• B. 18, 12
• C. 6, 4
• D. 0.6, 3.6

Answer 1

The correct option is A 1.8, 1.2

$1dyne={10}^{-5}$

${\mu }_0=4\pi \times {10}^{-7}Tm{A}^{-1}$

$m_1=3m$

$m_2=3m$

$F=\frac{{\mu }_0{m}_1\times m_2}{4\pi \times r^2}$ r=0.6m

$0.06\times {10}^{-5}=\frac{4\pi \times {10}^{-7}\times 3m\times 2m}{4\pi \times (0.6{)}^2}$

$m^2=(0.6{)}^2$

$m=0.6$

$m_1=1.8$

$m_2=1.2$