Question

The arrangement shown is placed in a vertical uniform magnetic field. The rods are of same length /and masses m. Initially they are placed in contact parallel to each other and pulled apart from rest such that net force (due to external as well as due to magnetic interaction) on rod at any instant is F. The current through resistor at an instant when rods are at separation 2s is $\frac{Bl}{R}\sqrt{\frac{\alpha FS}{m}}$. Find $\alpha$

Answer 1

8

Answer 2

The problem describes a system where two parallel conducting rods of length $l$ and mass $m$ are placed on conducting rails in a uniform magnetic field $B$. A resistor $R$ connects the rails. The rods are pulled apart from rest such that the net force on each rod at any instant is $F$. We need to find the value of $\alpha$ in the given expression for the current $I = \frac{Bl}{R}\sqrt{\frac{\alpha FS}{m}}$ when the separation between the rods is $2S$.

1. Net Force and Acceleration: The problem states that the net force on each rod at any instant is $F$. Since the mass of each rod is $m$, the acceleration of each rod is constant: $a = \frac{F}{m}$

2. Kinematics of the Rods: Let the rods start from rest at $t=0$ when their separation is zero. Due to the constant net force $F$ on each rod, they will accelerate away from each other. Let $x_1$ be the distance moved by the left rod and $x_2$ be the distance moved by the right rod. Since their accelerations are equal ($a = F/m$) and they start from rest, their velocities and displacements will be equal at any time $t$: $v_1 = v_2 = at = \left(\frac{F}{m}\right)t$ $x_1 = x_2 = \frac{1}{2}at^2 = \frac{1}{2}\left(\frac{F}{m}\right)t^2$

3. Total Separation and Relative Velocity: The total separation between the rods at time $t$ is $x = x_1 + x_2$. $x = \frac{1}{2}\left(\frac{F}{m}\right)t^2 + \frac{1}{2}\left(\frac{F}{m}\right)t^2 = \left(\frac{F}{m}\right)t^2$ The relative velocity (speed at which the separation increases) is $v_{rel} = v_1 + v_2$. $v_{rel} = \left(\frac{F}{m}\right)t + \left(\frac{F}{m}\right)t = 2\left(\frac{F}{m}\right)t$

4. Time to reach separation $2S$: We are interested in the instant when the separation is $2S$. Using the expression for total separation: $2S = \left(\frac{F}{m}\right)t^2$ Solving for $t^2$: $t^2 = \frac{2Sm}{F}$ So, $t = \sqrt{\frac{2Sm}{F}}$

5. Relative Velocity at separation $2S$: Substitute the value of $t$ into the expression for $v_{rel}$: $v_{rel} = 2\left(\frac{F}{m}\right)\sqrt{\frac{2Sm}{F}}$ To simplify, bring the $2(F/m)$ inside the square root: $v_{rel} = \sqrt{\left(2\frac{F}{m}\right)^2 \frac{2Sm}{F}} = \sqrt{4\frac{F^2}{m^2} \frac{2Sm}{F}} = \sqrt{\frac{8F^2Sm}{m^2F}} = \sqrt{\frac{8FS}{m}}$

6. Induced EMF and Current: The induced electromotive force (EMF) in the circuit due to the relative motion of the rods is given by: $\mathcal{E} = B l v_{rel}$ Substitute the expression for $v_{rel}$: $\mathcal{E} = B l \sqrt{\frac{8FS}{m}}$ The current flowing through the resistor $R$ is given by Ohm's law: $I = \frac{\mathcal{E}}{R} = \frac{B l}{R}\sqrt{\frac{8FS}{m}}$

7. Comparing with the given expression: The given expression for the current is $I = \frac{B l}{R}\sqrt{\frac{\alpha FS}{m}}$. Comparing our derived expression with the given one: $\frac{B l}{R}\sqrt{\frac{8FS}{m}} = \frac{B l}{R}\sqrt{\frac{\alpha FS}{m}}$ From this, we can clearly see that $\alpha = 8$.

The final answer is $\boxed{8}$.

Explanation of the solution:

1. Identify that the net force on each rod is constant ($F_{net} = F$).
2. Calculate the constant acceleration of each rod: $a = F/m$.
3. Determine the relative velocity ($v_{rel}$) and total separation ($x$) using constant acceleration kinematics: $v_{rel} = 2at$ and $x = at^2$.
4. Set the separation $x = 2S$ and solve for time $t$: $t = \sqrt{\frac{2Sm}{F}}$.
5. Substitute $t$ into the relative velocity expression to find $v_{rel}$ at separation $2S$: $v_{rel} = \sqrt{\frac{8FS}{m}}$.
6. Calculate the induced EMF: $\mathcal{E} = B l v_{rel}$.
7. Calculate the current: $I = \mathcal{E}/R = \frac{Bl}{R}\sqrt{\frac{8FS}{m}}$.
8. Compare this with the given current expression to find $\alpha = 8$.