Question

A conducting rod of length $l=1m$ moves in a magnetic field that varies with position as $B(x) = (2x)T$, where x is the distance from the origin. The rod was initially at $x=1m$ and moves to $x=2m$ in $2s$. The net resistance of circuit is $R = 0.5 \Omega$ then average current in the circuit is ______ ampere

Answer 1

3

Answer 2

The problem asks for the average current in a circuit where a conducting rod moves in a varying magnetic field.

1. Understand the Setup and Given Information:

• Length of the conducting rod, $l = 1 \text{ m}$.
• Magnetic field varies with position as $B(x) = (2x) \text{ T}$, where $x$ is the distance from the origin.
• Initial position of the rod, $x_1 = 1 \text{ m}$.
• Final position of the rod, $x_2 = 2 \text{ m}$.
• Time taken for the movement, $\Delta t = 2 \text{ s}$.
• Net resistance of the circuit, $R = 0.5 \Omega$.

2. Calculate the Magnetic Flux: The magnetic flux ($\Phi_B$) through the loop formed by the rod and the rails changes as the rod moves. Assuming the rails start at $x=0$, the area of the loop at any position $x$ is $A = l \cdot x$. However, since the magnetic field $B$ is not uniform but varies with $x$, we need to integrate to find the flux. The magnetic flux through the loop at a position $x$ is given by: $\Phi_B(x) = \int_0^x B(x') l dx'$ Substitute $B(x') = 2x'$ and $l=1$: $\Phi_B(x) = \int_0^x (2x') (1) dx' = \int_0^x 2x' dx'$ $\Phi_B(x) = \left[ (x')^2 \right]_0^x = x^2$

3. Calculate the Change in Magnetic Flux:

• Initial magnetic flux at $x_1 = 1 \text{ m}$: $\Phi_{B1} = (1)^2 = 1 \text{ Wb}$
• Final magnetic flux at $x_2 = 2 \text{ m}$: $\Phi_{B2} = (2)^2 = 4 \text{ Wb}$
• Change in magnetic flux, $\Delta\Phi_B$: $\Delta\Phi_B = \Phi_{B2} - \Phi_{B1} = 4 \text{ Wb} - 1 \text{ Wb} = 3 \text{ Wb}$

4. Calculate the Average Induced EMF: According to Faraday's Law of Induction, the average induced electromotive force (EMF) is the magnitude of the change in magnetic flux divided by the time taken: $\mathcal{E}_{avg} = \frac{|\Delta\Phi_B|}{\Delta t}$ Substitute the values: $\mathcal{E}_{avg} = \frac{3 \text{ Wb}}{2 \text{ s}} = 1.5 \text{ V}$

5. Calculate the Average Current: Using Ohm's Law, the average current ($I_{avg}$) in the circuit is the average induced EMF divided by the total resistance ($R$): $I_{avg} = \frac{\mathcal{E}_{avg}}{R}$ Substitute the values: $I_{avg} = \frac{1.5 \text{ V}}{0.5 \Omega} = 3 \text{ A}$

The average current in the circuit is 3 ampere.

The final answer is $\boxed{3}$.

Explanation of the solution:

1. Calculate the magnetic flux $\Phi_B(x)$ through the loop at any position $x$ by integrating $B(x')l dx'$ from $0$ to $x$. This gives $\Phi_B(x) = x^2$.
2. Determine the initial flux $\Phi_{B1}$ at $x=1m$ and final flux $\Phi_{B2}$ at $x=2m$. $\Phi_{B1} = 1^2 = 1$ Wb and $\Phi_{B2} = 2^2 = 4$ Wb.
3. Calculate the change in flux $\Delta\Phi_B = \Phi_{B2} - \Phi_{B1} = 4 - 1 = 3$ Wb.
4. Calculate the average induced EMF using Faraday's Law: $\mathcal{E}_{avg} = \frac{\Delta\Phi_B}{\Delta t} = \frac{3}{2} = 1.5$ V.
5. Calculate the average current using Ohm's Law: $I_{avg} = \frac{\mathcal{E}_{avg}}{R} = \frac{1.5}{0.5} = 3$ A.