Question

In a deflection magnetometer experiment in $tanA$ position, a short bar magnet placed at $18cm$ from the centre of the compass needle produces a deflection of ${30}^0$. If another magnet of same length, but $16$ times pole strength that of first magnet is placed in $tanB$ position at $36cm$, then the deflection is

• A. 30${}^0$
• B. 45${}^0$
• C. 60${}^0$
• D. 75${}^0$

Answer 1

Answer: (A)

30${}^0$

Given that,
$d_1=18cm$,
$d_2=36cm$,
$l_1=l_2$
$m_2=16m_1$,
we have,
For Tan A position
$\frac{2{\mu }_0{M}_1}{4\pi (d_1{)}^3}=B_H\tan {\theta }_1$
and for Tan B position
$\frac{{\mu }_0{M}_2}{4\pi (d_1{)}^3}=B_H\tan {\theta }_2$
$\Rightarrow \frac{2M_1}{M_2}=\frac{d_1^3\tan {\theta }_1}{d_2^3\tan {\theta }_2}$
But, M = ml, hence,
$\frac{2m_1}{16m_1}=\frac{d_1^{3}tan{\theta }_1}{d_2^{3}tan{\theta }_2}$
$\frac{tan{\theta }_2}{tan{\theta }_1}=\frac{8\times d_1^3}{d_2^3}$
$\frac{tan{\theta }_2}{tan{\theta }_1}=\frac{8(18{)}^3}{(36{)}^3}$
$\frac{tan{\theta }_2}{tan{\theta }_1}=1$
$tan{\theta }_2=tan{\theta }_1$
${\theta }_2={\theta }_1={30}^{\circ }$