First slide

Magnetic Dipole moment

Question

An $α - p a r t i c l e$ and a proton are moving in circular paths in the same uniform magnetic field with same kinetic energy. Then the ratio of magnetic moment due to the rotation of $α - p a r t i c l e$ to that of proton is

Difficult

A

1 : 4
B

4 : 1
C

2 : 1
D

1 : 1

Solution

(4)
$\begin{array}{l} \frac{1}{2} m v^{2} = K . E . = E \Rightarrow v = \sqrt{\frac{2 E}{m}} \\ A n g u l a r v e l o c i t y ω = \frac{Q B}{m} \\ ∴ r = R a d i u s o f c i r c u l a r p a t h \\ = \frac{v}{ω} = \frac{m}{Q B} . \sqrt{\frac{2 E}{m}} = \sqrt{2 m E} \frac{1}{Q B} \end{array}$
i = Equivalent current due to the moving charge.
$= \frac{Q ω}{2 π} = \frac{Q^{2} B}{2 π m}$
$∴ M = M a g n e t i c d i p o l e m o m e n t = π r^{2} \times i = \frac{π \times 2 E m}{Q^{2} B^{2}} \times \frac{Q^{2} B}{2 π m}$
$\begin{array}{l} \Rightarrow M = \frac{E}{B} \\ ∴ \frac{M_{α}}{M_{ρ}} = 1 \end{array}$

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