A metallic wire loop of side \( 0.1 \) m and resistance of \( 10 \Omega \) is moved with a constant velocity in a uniform magnetic field of \( 2 \text{ Wm}^{-2} \) as shown in the figure. The magnetic field is perpendicular to the plane of the loop. The loop is connected to a network of resistors. The velocity of loop so as to have a steady current of \( 1 \) mA in loop is:
Show Hint
In electromagnetic induction problems, use Faraday's Law to determine the induced emf and then apply Ohm's Law to find the required motion parameters. Ensure all units are consistent before calculations.
Step 1: Understanding Electromagnetic Induction From Faraday's Law of electromagnetic induction, the induced emf (\(\mathcal{E}\)) in the loop is given by: \[ \mathcal{E} = B l v \] where: - \( B = 2 \, \text{Wb/m}^2 \) (magnetic field strength), - \( l = 0.1 \, \text{m} \) (side length of the loop), - \( v \) (velocity of the loop, to be determined).
Step 2: Applying Ohm's Law The induced current \( I \) in the loop is given by: \[ I = \frac{\mathcal{E}}{R} \] where: - \( R = 10 \, \Omega \) (resistance of the loop), - \( I = 1 \) A (steady current in the loop).
Step 3: Solving for \( v \) Substituting the values: \[ 1 = \frac{(2 \times 0.1 \times v)}{10} \] \[ 1 = \frac{0.2 v}{10} \] \[ v = \frac{10}{0.2} = 2 \text{ m/s} = 2 \text{ cm/s} \]
