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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?

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

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

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