In the given figure a rod PQ of length 2.5 m can slide without friction. The value of R is \( 5.0\,\Omega \) and a magnetic field of 2 T is directed perpendicularly into the paper. The force F required to move the rod to the right at a constant speed of 1.0 m/s will be
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 5 N
The moving rod has a motional emf \( Blv \), which drives a current through the closed circuit with R. The field then exerts a force \( BIl \) on the current in the rod, which by Lenz's law opposes the motion. At constant speed the net force is zero, so the applied force F equals this magnetic force.
\( B = 2\,\mathrm{T} \)
\( l = 2.5\,\mathrm{m} \)
\( v = 1.0\,\mathrm{m\,s^{-1}} \)
\( R = 5.0\,\Omega \)
\( \displaystyle \varepsilon = Blv = 2 \times 2.5 \times 1.0 = 5\,\mathrm{V} \)
\( \displaystyle I = \frac{\varepsilon}{R} = \frac{5}{5.0} = 1\,\mathrm{A} \)
\( \displaystyle F = BIl = 2 \times 1 \times 2.5 = 5\,\mathrm{N} \)
This is the same as \( F = B^{2}l^{2}v/R \).