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Q23PE

Expert-verifiedFound in: Page 861

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**Derive an expression for the current in a system like that in figure below, under the following conditions. The resistance between the rails is R , the rails and the moving rod are identical in cross section A and have the same resistivity ρ. The distance between the rails is l, and the rod moves at constant speed v perpendicular to the uniform field B. At time zero, the moving rod is next to the resistance R.**

The expression is $I\left(t\right)=\frac{ABLv}{AR+\rho \left(L+vt\right)}$

**According to Lenz, the direction of induced emf and hence the direction of the induced current is such that it will oppose the motion of the conductor in the external field.**

The current (I) can be calculated using Ohm's law, as-

$I\left(t\right)=\frac{\epsilon \left(t\right)}{R\left(t\right)}$

Where $\left(\epsilon \right)$ is emf and R is Resistance.

The emf $\left(\epsilon \right)$ will be given as

$\epsilon \left(t\right)=BLv$

The resistance (R) of the metal structure, which means,

$R\left(t\right)=R+\rho \left(\frac{L+vt}{A}\right)$

Combining the two expressions, we get,

$I\left(t\right)=\frac{BLv}{R+\rho \left(\frac{L+vt}{A}\right)}$

The above equation can further be expressed as-

$I\left(t\right)=\frac{ABLv}{AR+\rho \left(L+vt\right)}$

**This is the required expression.**

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