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?

  • 1
    $\begingroup$ What have I done wrong? In general, check-my-work questions are off-topic here. $\endgroup$
    – G. Smith
    Sep 20, 2020 at 5:45

1 Answer 1


According to the flux approach,


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:



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.