Question

A 1 m long metal rod AB completes the circuit as shown in figure. The area of circuit is perpendicular to the magnetic field of 0.10 T. If the resistance of the total circuit is 2 \(\Omega\) then the force needed to move the rod towards right with constant speed (v) of 1.5 m/s is _____ N.

• A. \(5.7 \times 10^{-2}\)
• B. \(7.5 \times 10^{-3}\)
• C. \(7.5 \times 10^{-2}\)
• D. \(5.7 \times 10^{-3}\)

Hint:

An alternative approach is to use the principle of energy conservation. The mechanical power supplied by the external force must equal the electrical power dissipated as heat in the resistor.
Mechanical Power \(P_{\text{mech}} = F_{\text{app}} \cdot v\).
Electrical Power \(P_{\text{elec}} = I^2 R = \frac{\mathcal{E}^2}{R} = \frac{(BLv)^2}{R}\).
Equating them: \(F_{\text{app}} v = \frac{B^2 L^2 v^2}{R} \implies F_{\text{app}} = \frac{B^2 L^2 v}{R}\). This gives the answer in one step.

Answer 1

The Correct Option is B

The problem involves a conducting metal rod moving with constant velocity in a uniform magnetic field. Due to electromagnetic induction, an emf is induced in the rod, producing a current and hence a magnetic force. We must determine the external force required to maintain constant speed.

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Step-by-Step Solution

1. Induced emf

   The induced emf in a rod of length \(l\) moving with speed \(v\) perpendicular to a magnetic field \(B\) is:

   \[ \varepsilon = B l v \]

   Substituting given values:

   \[ \varepsilon = 0.10 \times 1 \times 1.5 = 0.15~\text{V} \]

2. Current in the circuit

   Using Ohm’s law:

   \[ I = \frac{\varepsilon}{R} \]

   With \(R = 2~\Omega\):

   \[ I = \frac{0.15}{2} = 0.075~\text{A} \]

3. Magnetic force on the rod

   The magnetic force acting on a current-carrying conductor is:

   \[ F = B I l \]

   Substituting the values:

   \[ F = 0.10 \times 0.075 \times 1 = 0.0075~\text{N} \]

4. To maintain constant velocity, an external force equal in magnitude and opposite in direction to this magnetic force must be applied.

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Final Answer:

\(\boxed{7.5 \times 10^{-3}~\text{N}}\)