Question
A thin uniform rod, length ℓ = 0.200 m with negligible mass, is attached to the floor by a frictionless hinge at a fixed point P. A horizontal spring connects the other end of the rod to a vertical wall. The rod is in a uniform magnetic field B = 0.500 tesla directed into the plane of paper. There is a current i = 10.0 A in the rod in the direction shown. Force constant of the spring is 5 N/m. The rod is in equilibrium at θ = tan−1(4/3).
Statement I: Torque on the rod due to magnetic force is 0.1 N m clockwise.
Statement II: In equilibrium the energy stored in the spring is 0.039 J.
Options
Solution
[Key Concept: The magnetic force on a straight current-carrying rod in a uniform field acts at its midpoint; in equilibrium the spring torque about the hinge balances it.]
(a)Magnetic force on rod
$$F_{m} = I(\overrightarrow{\mathcal{l}} \times \overrightarrow{B}) = 10 \times 0.2 \times 0.5sin90^{\circ} = 1$$ Newton
Torque due to magnetic field
$= (Fm)$ at centre of mass of rod $\times r_{\bot}$
$${= 1 \times \frac{L}{2} = 1 \times \frac{0.2}{2} }{= 0.1Nm }$$clockwise
(b)Toque due to spring
$$\begin{matrix} & T_{\text{spring~}} = \left( F_{s} \right)Lsin\theta \\ & \ \end{matrix}$$
$$= kx \times 0.2 \times \frac{4}{5}$$
Torque of spring must become torque due to magnetic field
$$kx \times 0.2 \times \frac{4}{5} = 0.1$$
$\begin{matrix} & x = \frac{0.1 \times 5}{4 \times 0.2 \times 5} = \frac{1}{8} = 0.125\text{ }m \\ & \ \therefore\text{~Energy stored in spring~} = \frac{1}{2}{kx}^{2} \\ & \ = \frac{1}{2} \times 5 \times (0.125)^{2} \\ & \ = 0.039\text{~Joule~} \end{matrix}$
d option is correct
Both statement (1) and (2) are correct
Create a free account to view solution
View Solution Free