A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by 60° is W. Now the torque required to keep the magnet in this new position is √k W. Find k.

Question

A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by 60° is W. Now the torque required to keep the magnet in this new position is √k W. Find k. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

τ=M B sin 60° ....(i) W=M B(1- cos 60°)....(ii) From (i) and (ii), we get (τ/W)=(√3 / 2/1 / 2) ⇒ τ=W √3

Q. A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by $6 0^{\circ}$ is $W$ . Now the torque required to keep the magnet in this new position is $k W$ . Find $k$ .

$τ = MB sin 6 0^{\circ}$ ....(i)
$W = MB (1 - cos 6 0^{\circ})$ ....(ii)
From (i) and (ii), we get
$\frac{τ}{W} = \frac{3 /2}{1/2}$
$\Rightarrow τ = W 3$

Q. A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by 60∘ is W. Now the torque required to keep the magnet in this new position is k​W. Find k.

τ=MBsin60∘ ....(i) W=MB(1−cos60∘)....(ii) From (i) and (ii), we get Wτ​=1/23​/2​ ⇒τ=W3​

Q. A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by $6 0^{\circ}$ is $W$ . Now the torque required to keep the magnet in this new position is $k W$ . Find $k$ .

$τ = MB sin 6 0^{\circ}$ ....(i)
$W = MB (1 - cos 6 0^{\circ})$ ....(ii)
From (i) and (ii), we get
$\frac{τ}{W} = \frac{3 /2}{1/2}$
$\Rightarrow τ = W 3$