An electron in the ground state of hydrogen atom is revolving in anticlockwise direction in a circular orbit of radius R. The atom is placed in a uniform magnetic induction B→ such that the normal to the plane of electron’s orbit makes an angle of 300 with the magnetic induction. Find the torque experienced by the orbiting electron.

Magnetic moment (M)=iA;|τ→|=MBsin⁡θML=e2m⇒M=e2mnh2π|T→|=MBsinθ=e2mnh2πB12=ehB8πm n=1

An electron in the ground state of hydrogen atom is revolving in anticlockwise direction in a circular orbit of radius R. The atom is placed in a uniform magnetic induction B→ such that the normal to the plane of electron’s orbit makes an angle of 300 with the magnetic induction. Find the torque experienced by the orbiting electron.

Magnetic moment (M)=iA;|τ→|=MBsin⁡θML=e2m⇒M=e2mnh2π|T→|=MBsinθ=e2mnh2πB12=ehB8πm n=1

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An electron in the ground state of hydrogen atom is revolving in anticlockwise direction in a circular orbit of radius R. The atom is placed in a uniform magnetic induction $\vec{B}$ such that the normal to the plane of electron’s orbit makes an angle of 30⁰ with the magnetic induction. Find the torque experienced by the orbiting electron.

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$\frac{8 eh}{πmB}$

$\frac{Beh}{πm}$

$\frac{eh}{4 πm}$

$\frac{ehB}{8 πm}$

answer is D.

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Detailed Solution

Magnetic moment $(M) = iA; | \vec{τ} | = MBsin θ$

$\frac{M}{L} = \frac{e}{2 m}$

$\Rightarrow M = \frac{e}{2 m} (\frac{n h}{2 π})$

$| \vec{T} | = M B \sin θ = \frac{e}{2 m} (\frac{n h}{2 π}) B \frac{1}{2} = \frac{ehB}{8 π m} (n = 1)$

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