Consider two parallel conducting frictionless rails in a gravity free rails parallel to x axis. A movable conductor PQ( y direction) of length $$l$$ slides on those rails. The rails are also connected by a fixed wire AB with a resistor of resistance $$R$$. Suppose a magnetic field exists in region which varies as $$B = cx$$The magnetic field is perpendicular to the plane of the system. Initially PQ is given some velocity $$v_0$$ in the x direction. Let the velocity at any instant be $$v$$ and the distance from AB be $$x$$

1. According to the flux approach, $$\Phi=cx^2l$$ $$\frac{d\Phi}{dt}=2cxlv$$ Force on conductor $$= 2c^2x^2l^2v$$

2. According to motional EMF approach $$\epsilon = cxvl$$ Force on conductor $$= c^2x^2l^2v$$

What have I done wrong?

• What have I done wrong? In general, check-my-work questions are off-topic here. – G. Smith Sep 20 at 5:45

According to the flux approach,

Φ=𝑐𝑥2𝑙

This step is incorrect. If I take any dx element at a distance x from the AB, then area of element is $$ldx$$ and magnetic field $$B=cx\tag1$$.

Then Flux $$\phi$$ is given by: $$d\phi = B dA = cx l dx$$ Integrating the expression:

$$=>\phi = \int cl xdx$$from x=0 to x=x, we get: $$\phi = \frac12 clx^2$$ EMF $$\epsilon$$ is given by: $$\epsilon=\frac{d\phi}{dt}=clx\frac{dx}{dt}=clxv\tag2$$

Further force on conductor is: $$F=ilB$$ where $$i=\frac{\epsilon}{R}\tag3$$

Substituting the known expressions from eq(1),eq(2) and eq(3) at position x:

$$F=\frac{c^2L^2x^2v}{R}$$