The work required to rotate a magnetic needle by 60 $^{0}$ from equilibrium position in a uniform magnetic field is W. The torque required to hold it in that position is

W
$\sqrt{3} W$
$\frac{W}{2}$
$\frac{\sqrt{3}}{2}$ W

Question

The work required to rotate a magnetic needle by 60-0- from equilibrium position in a uniform magnetic field is W- The torque required to hold it in that position isWsqrt-3-Wdfrac-W-2-dfrac-sqrt-3-2-W

Toppr Admin · Accepted Answer

-x3B8-1-0-x3B8-2-60U1-x2212-mBcos0-x2212-mBU2-x2212-mBcos600-x2212-mBzw-U2-x2212-U1-mBzz-mBsin-x3B8-mBsin60-mBz-x221A-3z-w-x221A-3

The work required to rotate a magnetic needle by 600 from equilibrium position in a uniform magnetic field is W. The torque required to hold it in that position isW√3WW2√32W

θ1=0θ2=60U1=−mBcos0=−mBU2=−mBcos600=−mBzw=U2−U1=mBzz=mBsinθ=mBsin60=mBz√3z=w√3

The work required to rotate a magnetic needle by 60 $^{0}$ from equilibrium position in a uniform magnetic field is W. The torque required to hold it in that position is

W
$\sqrt{3} W$
$\frac{W}{2}$
$\frac{\sqrt{3}}{2}$ W

$θ_{1} = 0$
$θ_{2} = 60$
$U_{1} = - m B c o s 0$
$= - m B$
$U_{2} = - m B c o s 60^{0}$
$= - \frac{m B}{z}$
$w = U_{2} - U_{1}$
$= \frac{m B}{z}$
$z = m B s i n θ$
$= m B s i n 60$
$= \frac{m B}{z} \sqrt{3}$
$z = w \sqrt{3}$