Contradiction in Faraday's law and Motional EMF

Question

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 $x$

According to the flux approach, $$Phi=cx^2l$$
$$frac{dPhi}{dt}=2cxlv$$
Force on conductor $= 2c^2x^2l^2v$
According to motional EMF approach
$$epsilon = cxvl$$
Force on conductor $= c^2x^2l^2v$

What have I done wrong?

electromagnetic induction electromagnetism forces homework and exercises magnetic fields

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

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

What have I done wrong?

Tony Stark · Accepted Answer

According to the flux approach,

Φ=??2?

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 $$B=cxtag1$$.
Then Flux $phi$ is given by:
$$dphi = B dA = cx l dx$$
Integrating the expression:
$$=>phi = int cl xdx$$from  x=0 to x=x, we get:
$$phi = frac12 clx^2$$
EMF $epsilon$ is given by:
$$epsilon=frac{dphi}{dt}=clxfrac{dx}{dt}=clxvtag2$$

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

Contradiction in Faraday's law and Motional EMF

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