A rod of mass textm and length 2 textl is connected to two massless strings as shown in figure. A wooden disc of the same mass textm is fixed at the mid-point of the rod in such a way that the plane of the disc is perpendicular to the rod. A massless wire is wrapped around the disc and it is carrying a current I . There is vertically upward magnetic field textB at the location of disk. (Sense of current in the loop when seen from string P side is clockwise) The tension in the string textP is

Question

A rod of mass textm and length 2 textl is connected to two massless strings as shown in figure. A wooden disc of the same mass textm is fixed at the mid-point of the rod in such a way that the plane of the disc is perpendicular to the rod. A massless wire is wrapped around the disc and it is carrying a current I . There is vertically upward magnetic field textB at the location of disk. (Sense of current in the loop when seen from string P side is clockwise) <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-iwjejx0ifskv.png" /> The tension in the string textP is (A) mg+(I π R2 B/2 l) (B) mg-(I π R2 B/2 l) (C) mg (D) 2mg

Tardigrade Admin · Accepted Answer

net torque about the disc is zero TP× l=TQ× l+Iπ R2B and TP+TQ=2mg Using the above equations we get TP=mg+(Iπ R2 B/2 l)

$m g + \frac{I π R ^{2} B}{2 l}$

$m g - \frac{I π R ^{2} B}{2 l}$

$m g$

$2 m g$

net torque about the disc is zero
$T_{P} \times l = T_{Q} \times l + I π R^{2} B$
and $T_{P} + T_{Q} = 2 m g$
Using the above equations we get
$T_{P} = m g + \frac{I π R ^{2} B}{2 l}$

mg+2lIπR2B​

mg−2lIπR2B​

mg

2mg

net torque about the disc is zero TP​×l=TQ​×l+IπR2B and TP​+TQ​=2mg Using the above equations we get TP​=mg+2lIπR2B​

$m g + \frac{I π R ^{2} B}{2 l}$

$m g - \frac{I π R ^{2} B}{2 l}$

$m g$

$2 m g$

net torque about the disc is zero
$T_{P} \times l = T_{Q} \times l + I π R^{2} B$
and $T_{P} + T_{Q} = 2 m g$
Using the above equations we get
$T_{P} = m g + \frac{I π R ^{2} B}{2 l}$