Question 93: Magnetism and Matter

Question

At some location on earth, the horizontal component of earth’s magnetic field is 18 × 10–6 T. At this location, magnetic needle of length 0.12 m and pole strength 1.8 Am is suspended from its mid-point using a thread, it makes 45o angle with horizontal in equilibrium. To keep this needle horizontal, the vertical force that should be applied at one of its ends is -

Accepted Answer

Without applied forces, (in equilibrium position) the needle will stay in the resultant magnetic field of earth. Hence, the dip ' $	heta$ ' at this place is $45^{\circ}$ (given).
We know that, horizontal and vertical components of earth's magnetic field $\left(B_Hight.$ and $B_V$ ) are related as
$$
\frac{B_V}{B_H}=	an 	heta
$$
Here, $	heta=45^{\circ}$ and $B_H=18 	imes 10^{-6} \mathrm{~T}$
$$

\Rightarrow  B_V=B_H 	an 45^{\circ} $$
$$
\Rightarrow  B_V=B_H=18 	imes 10^{-6} \mathrm{~T} \quad\left(\because 	an 45^{\circ}=1ight)$$
Now, when the external force $F$ is applied, so as to keep the needle stays in horizontal position is shown below,
Taking torque at point $P$, we get
$$
\begin{array}{ll} 
 m B_V 	imes 2 l=F l \\
	herefore F=2 	imes m B_V
\end{array}
$$
Substituting the given values, we get
$$
\begin{aligned}
& =2 	imes 1.8 	imes 18 	imes 10^{-6} \\
& =6.48 	imes 10^{-5}=6.5 	imes 10^{-5} \mathrm{~N}
\end{aligned}
$$

Answer

3.6 $$ 	imes $$ 10$$-$$5 N

Answer

1.8 $$ 	imes $$ 10$$-$$5 N

Answer

1.3 $$ 	imes $$ 10$$-$$5 N

Answer

6.5 $$ 	imes $$ 10$$-$$5 N

Magnetism and Matter

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