Question

A 60 V DC source with an internal resistance \(R_{int} = 0.5 \, \Omega\) is connected through a switch to a pair of infinitely long rails separated by \(l = 1\) m as shown in the figure. The rails are placed in a constant, uniform magnetic field of flux density \(B = 0.5\) T, directed into the page. A conducting bar placed on these rails is free to move. At the instant of closing the switch, the force induced on the bar is

 

• A. 60 N in the direction of $\hat{x}$
• B. 60 N opposite to the direction of $\hat{x}$
• C. 120 N in the direction of $\hat{x}$
• D. 120 N opposite to the direction of $\hat{x}$

Hint:

Remember the Lorentz force law $\mathbf{F} = I (\mathbf{l} \times \mathbf{B})$ for the force on a current-carrying conductor in a magnetic field. Use the right-hand rule to correctly determine the direction of the force based on the directions of the current and the magnetic field.

Answer 1

The Correct Option is A

Step 1: Determine the initial current in the conducting bar.
At the instant the switch is closed, the bar is stationary, so there is no induced back EMF. The current in the circuit is determined by the voltage source and the total resistance in the loop. The total resistance is the internal resistance of the source $R_{int} = 0.5 \Omega$ (since the resistances of the rails and the bar are zero).
The current $I$ flowing through the bar is:
$$I = \frac{V}{R_{int}} = \frac{60 { V}}{0.5 \Omega} = 120 { A}$$ The direction of this current in the conducting bar is upwards (from the lower rail to the upper rail).
Step 2: Calculate the magnetic force on the conducting bar using the Lorentz force law.
The magnetic force $\mathbf{F}$ on a current-carrying conductor in a magnetic field $\mathbf{B}$ is given by $\mathbf{F} = I (\mathbf{l} \times \mathbf{B})$, where $\mathbf{l}$ is the vector representing the length and direction of the current in the conductor.
Here, the magnitude of the current is $I = 120$ A, the length of the bar is $l = 1$ m, and the magnetic field strength is $B = 0.5$ T, directed into the page. The direction of the current $\mathbf{l}$ is upwards. Using the right-hand rule for the cross product or the Lorentz force on a current-carrying wire: point your fingers in the direction of the current (upwards), curl them into the direction of the magnetic field (into the page). Your thumb will point in the direction of the force. In this case, the force is directed to the right, which is the direction of $\hat{x}$. The magnitude of the force is:
$$F = I l B = (120 { A}) (1 { m}) (0.5 { T}) = 60 { N}$$ The direction of the force is along $\hat{x}$.