A rod of length 2 m slides with a speed of 5 ms-1 on a rectangular conducting frame as shown in figure. There exists a uniform magnetic field of 0.04 T perpendicular to the plane of the figure. If the resistance of the rod is 3 Ω. The current through the rod is
The induced electromotive force (EMF) in the rod is given by the formula: \[ \text{EMF} = B \times v \times l \] Where: \( B \) is the magnetic field strength (0.04 T), \( v \) is the velocity of the rod (5 ms\(^{-1}\)), \( l \) is the length of the rod (2 m). Substituting the values: \[ \text{EMF} = 0.04 \times 5 \times 2 = 0.4 \, \text{V} \] Now, using Ohm’s Law, the current through the rod is: \[ I = \frac{\text{EMF}}{R} = \frac{0.4}{3} = 0.1333 \, \text{A} = 133 \, \text{mA} \] Thus, the current through the rod is 133 mA.
Was this answer helpful?
1
0
Hide Solution
Verified By Collegedunia
Approach Solution -2
The induced EMF \( \mathcal{E} \) in the moving rod can be calculated using the formula for motional EMF: \[ \mathcal{E} = B \cdot L \cdot v \] where: \( B \) is the magnetic field strength (\( 0.04 \, \text{T} \)), \( L \) is the length of the rod (\( 2 \, \text{m} \)), \( v \) is the velocity of the rod (\( 5 \, \text{ms}^{-1} \)).
Substituting the given values: \[ \mathcal{E} = 0.04 \cdot 2 \cdot 5 = 0.4 \, \text{V} \] Now, the current \( I \) through the rod can be found using Ohm's law: \[ I = \frac{\mathcal{E}}{R} \] where \( R \) is the resistance of the rod (\( 3 \, \Omega \)). Substituting the values: \[ I = \frac{0.4}{3} = 0.133 \, \text{A} = 133 \, \text{mA} \] Thus, the current through the rod is \( 133 \, \text{mA} \).
