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 released, it starts performing simple harmonic motion of time period T. If the mass of the loop is m then :

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 released, it starts performing simple harmonic motion of time period T. If the mass of the loop is m then : (A) T=√(π m/IB) (B) T=√(π m/2IB) (C) T=√(2m/IB) (D) T=√(2π m/IB)

Tardigrade Admin · Accepted Answer

vecT = vecM × vecB = -MBsinθ Iα = -MB sinθ for small θ, α = -(MB/I) θ ω = √(MB/I) = √((i)(π R2)B/((mR2)2)) ω = √(2iπ B/m) ∴ = T = (2π/ω) = √(2iπ B/iB)

$T = \frac{πm}{I B}$

$T = \frac{πm}{2 I B}$

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

$T = \frac{2 πm}{I B}$

$T = M \times B = - MB s in θ$
$I α = - MB s in θ$
for small $θ$ ,
$α = - \frac{MB}{I} θ$
$ω = \frac{MB}{I} = \frac{( i ) ( π R ^{2} ) B}{( \frac{m R ^{2}}{2} )}$
$ω = \frac{2 iπ B}{m}$
$∴ = T = \frac{2 π}{ω} = \frac{2 iπ B}{i B}$

T=IBπm​​

T=2IBπm​​

T=IB2m​​

T=IB2πm​​

T=M×B=−MBsinθ Iα=−MBsinθ for small θ, α=−IMB​θ ω=IMB​​=(2mR2​)(i)(πR2)B​​ ω=m2iπB​​ ∴=T=ω2π​=iB2iπB​​

$T = \frac{πm}{I B}$

$T = \frac{πm}{2 I B}$

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

$T = \frac{2 πm}{I B}$

$T = M \times B = - MB s in θ$
$I α = - MB s in θ$
for small $θ$ ,
$α = - \frac{MB}{I} θ$
$ω = \frac{MB}{I} = \frac{( i ) ( π R ^{2} ) B}{( \frac{m R ^{2}}{2} )}$
$ω = \frac{2 iπ B}{m}$
$∴ = T = \frac{2 π}{ω} = \frac{2 iπ B}{i B}$