Question

A metallic rod of length L = 2m and mass M = 1 kg is moving under the action of two unequal forces F₁ = 5N and F₂ = 2N (directed opposite to each other) acting at its end along its length. Ignore gravity and any other external magnetic field. Specific charge of electrons is $\frac{e}{m}$. The potential difference between the ends of the rod in steady state is $V = \frac{km}{e}$(Sl units). Find the value of k.

Answer 1

6

Answer 2

The problem describes a metallic rod accelerating under the action of two unequal forces. Due to this acceleration, an electric field is established inside the rod, leading to a potential difference across its ends.

1. Calculate the acceleration of the rod: The net force acting on the rod is $F_{net} = F_1 - F_2$. Given $F_1 = 5N$ and $F_2 = 2N$, so $F_{net} = 5N - 2N = 3N$. The mass of the rod is $M = 1kg$. Using Newton's second law, the acceleration of the rod is: $a = \frac{F_{net}}{M} = \frac{3N}{1kg} = 3 \text{ m/s}^2$

2. Determine the electric field inside the rod: When the metallic rod accelerates, the free electrons inside it, due to their inertia, tend to lag behind the positive ions. This causes a separation of charge, leading to the establishment of an internal electric field. In steady state, the free electrons must also accelerate with the rod. The force required to accelerate an electron (mass $m$, charge $e$) is $ma$. This force is provided by the electric field $E$. The force on an electron due to the electric field is $F_E = eE$. For the electron to accelerate with the rod, the net force on it must be $ma$: $eE = ma$ Therefore, the magnitude of the electric field inside the rod is: $E = \frac{ma}{e}$

3. Calculate the potential difference across the ends of the rod: The potential difference $V$ across the ends of the rod is given by $V = EL$, where $L$ is the length of the rod. Given $L = 2m$. Substitute the expression for $E$: $V = \left(\frac{ma}{e}\right)L$ Now, substitute the values of $a$ and $L$: $V = \frac{m(3 \text{ m/s}^2)(2 \text{ m})}{e}$ $V = \frac{6m}{e}$

4. Compare with the given form and find k: The problem states that the potential difference is $V = \frac{km}{e}$. Comparing our derived expression $V = \frac{6m}{e}$ with the given form, we find that: $k = 6$

The value of k is 6.

Explanation of the solution:

1. Calculate the acceleration of the rod using Newton's second law: $a = (F_1 - F_2) / M$.
2. In steady state, the electrons inside the rod accelerate with the rod. The force causing this acceleration is due to the internal electric field $E$. So, $eE = ma$.
3. The potential difference across the rod is $V = EL$.
4. Substitute $E = ma/e$ into the potential difference equation: $V = (ma/e)L$.
5. Substitute the calculated values of $a$ and given $L$ to find $V = \frac{6m}{e}$.
6. Compare this with the given form $V = \frac{km}{e}$ to find $k=6$.