First slide

Magnetism

Question

A magnetic needle of pole strength $20 \sqrt{3}$ Am is pivoted at its centre. Its North pole is pulled eastward by a string. The horizontal force required to produce a deflection of $30^{0}$ from magnetic meridian (take $B_{H} = 10^{- 4} T$ )

Moderate

A

$4 \times 10^{- 3} N$
B

$2 \times 10^{- 3} N$
C

$\frac{2}{\sqrt{3}} \times 10^{- 3} N$
D

$4 \sqrt{3} \times 10^{- 3} N$

Solution

from the figure    $\cos θ = \frac{y}{l}, y = l \cos θ$
$\begin{array}{l} T o r q u e d u e t o F i s b a l a n c e d b y t o r q u e d u e t o B_{H} . \\ \Rightarrow T_{tension} = T_{B_{H}} \\ \Rightarrow F y = M B_{H} \sin θ \\ \Rightarrow F l \cos θ = m (2 l) B_{H} \sin θ \\ \Rightarrow F = 2 m B_{H} \tan θ \end{array}$
Given $m = 20 \sqrt{3} A m$
   $\begin{array}{l} B_{H} = 10^{- 4} T \\ θ = 30 ° \\ U \sin g, F = 2 m B_{H} \tan θ \\ \Rightarrow F = 2 \times 20 \sqrt{3} \times 10^{- 4} \times \frac{1}{\sqrt{3}} = 4 \times 10^{- 3} N \end{array}$

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