First slide

Applications involving charged particles moving in a magnetic field

Question

A small circular loop of conducting wire has radius a and carries current I. It is placed in a uniform magnetic field B perpendicular to its plane such that when rotated slightly about its diameter and simple harmonic motion of time period T. If the mass of the loop is m then

Moderate

A

$T= \sqrt{\frac{π m}{I B}}$
B

$T = \sqrt{\frac{π m}{2 I B}}$
C

$T = \sqrt{\frac{2 m}{I B}}$
D

$T = \sqrt{\frac{2 π m}{I B}}$

Solution

$\bar{τ} = \bar{M} \times \bar{B}$
The deflecting torque ${\bar{τ}}_{d} = I α$ , Restoring torque ${\bar{τ}}_{r} = - M B \sin θ$
At equilibrium ${\bar{τ}}_{d} = {\bar{τ}}_{r}$
$\begin{array}{l} I α = - M B \sin θ \\ \Rightarrow I α = - M B θ [f o r s m a l l θ, \sin θ ≃ θ] \\ \Rightarrow α = - \frac{M B}{I} θ \\ \Rightarrow ω = \sqrt{\frac{M B}{I}} \end{array}$
$∴ T = \frac{2 π}{ω} = 2 π \sqrt{\frac{I}{M B}}$
We know, $M = n I A$ (Magnetic moment)
$\begin{array}{l} \Rightarrow M = 1 \times I \times π R^{2} a n d \\ M o m e n t o f I n e r t i a = I = \frac{m R^{2}}{2} \\ T h u s, T = 2 π \sqrt{\frac{\frac{m R^{2}}{2}}{I . π R^{2} B}} \end{array}$
$= \sqrt{\frac{4 π^{2} . m R^{2}}{2. I . π R^{2} B}} = \sqrt{\frac{2 π m}{I B}}$

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