In the given figure a rod PQ of length 2.5 m can slide without friction. The value of R is 5.0 Ω 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

5 N

The correct answer is Option (2) → 5 N

Given:

Length of rod, $l = 2.5\ \text{m}$

Magnetic field, $B = 2\ \text{T}$

Resistance, $R = 5\ \Omega$

Velocity of rod, $v = 1.0\ \text{m/s}$

Induced emf:

$ \varepsilon = B\,l\,v $

$ \varepsilon = 2 \times 2.5 \times 1 = 5\ \text{V}$

Induced current:

$ I = \frac{\varepsilon}{R} = \frac{5}{5} = 1\ \text{A}$

Magnetic force on the rod:

$ F = B\,I\,l $

$ F = 2 \times 1 \times 2.5 = 5\ \text{N}$

Therefore, the force required to move the rod at constant speed is

$F = 5\ \text{N}$