JEE MAIN - Physics (2025 - 28th January Evening Shift - No. 25)
_28th_January_Evening_Shift_en_25_1.png)
A conducting bar moves on two conducting rails as shown in the figure. A constant magnetic field B exists into the page. The bar starts to move from the vertex at time t = 0 with a constant velocity. If the induced EMF is E ∝ tn, then value of n is _________.
Answer
1
Explanation
_28th_January_Evening_Shift_en_25_2.png)
$$\mathrm{E}=\ell \mathrm{vB}$$
$E=\frac{2 x}{\sqrt{3}} \times v B$ and $x=v t$
$$\mathrm{E}=\frac{2}{\sqrt{3}} \mathrm{v}^2 \mathrm{Bt}$$
$E \propto t^1$
Comments (0)
