First slide

Magnetism

Question

A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through $60^{0}$ .What is the torque needed to maintain the needle in that position?

Moderate

A

$w$
B

$\frac{\sqrt{3}}{2} w$
C

$\sqrt{3} w$
D

$\frac{w}{2}$

Solution

In the case of a dipole in a magnetic field.
$W = M B (\cos θ_{1} - \cos θ_{2})$
and $Γ = M B \sin θ$
here $θ_{1} = 0^{0} θ_{2} = 60 °$
$W = M B (1 - \cos θ)$
$\begin{array}{l} \Rightarrow W = M B 2 \sin^{2} θ / 2 (∵ (1 - \cos θ) = 2 \sin^{2} θ / 2) \\ D i v i d i n g, \frac{Γ}{W} = \frac{M B \sin θ}{M B 2 \sin^{2} θ / 2} = \frac{2 \sin \frac{θ}{2} \cos \frac{θ}{2}}{2 \sin^{2} θ / 2} \end{array}$
$\begin{array}{l} \Rightarrow \frac{Γ}{W} = \cot \frac{θ}{2} \\ \Rightarrow Γ = W \cot 30 ° \\ \Rightarrow Γ = \sqrt{3} W \end{array}$

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