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. lf the resistance of the total circuit is 2Ω then the force needed to move the rod towards right with constant speed (v) of 1.5 m/s is ___ N

The motional EMF ($$\varepsilon$$) induced in the moving rod is given by:

$$\varepsilon = B l v$$

The induced current ($$I$$) flowing through the circuit is:

$$I = \dfrac{\varepsilon}{R} = \dfrac{B l v}{R}$$

The magnetic force ($$F_m$$) acting on the current-carrying rod opposes its motion (Lenz's Law) and is given by:

$$F_m = I(l\times B)$$

To keep the rod moving at a constant speed, the external applied force ($$F$$) must exactly balance this magnetic force ($$F = F_m$$):

the direction of force is towards left(use right hand thumb rule )

$$F = \left( \dfrac{B l v}{R} \right) l B$$

$$F = \dfrac{B^2 l^2 v}{R}$$

Substitute the values into the equation:

$$F = \dfrac{(0.10)^2 \times (1)^2 \times 1.5}{2}$$

$$F = \dfrac{0.015}{2}$$

$$F = 7.5 \times 10^{-3} \text{ N}$$