Question

A conducting slider of resistance $R(10\Omega )$ mass $50g$ & length $10cm$ is kept on a U – shaped frame as shown in figure. There is uniform magnetic field$(B=0.1T)$ perpendicular to plane of frame. These slider is attached to a spring$(K=0.5\raisebox{1ex}{$N$}\!\left/ \!\raisebox{-1ex}{$m$}\right.)$. It is displaced by $A_{Ɵ}$ & released time is which amplitude become $\raisebox{1ex}{$A$}\!\left/ \!\raisebox{-1ex}{$e$}\right.$ is

• A. $9000s$
• B. $10000s$
• C. $12000s$
• D. $15000s$

Answer 1

The correct option is B

$10000s$

Step 1. Given Data:

$R=10\Omega$

Mass $=50g$

Length $=10cm$

$B=0.1T$

$K=0.5N{m}^{-1}$

Step 2. As we know that:

$$\begin{gathered} \begin{array}{l}-Kx-i\ell B=m\frac{d^{2}x}{d{\ell }^2}\end{array} \\ \Rightarrow \begin{array}{l}-Kx-\frac{B^2{\ell }^2}{R}\frac{dx}{d\ell }-m\frac{d^{2}x}{d{\ell }^2}=0\end{array} \end{gathered}$$

Step 3. By comparing with:

$\begin{array}{l}-Kx-b\frac{dx}{dt}-m\frac{d^{2}x}{d{\ell }^2}=0\end{array}$

$A=A_0{e}^{\left(\frac{-b}{2m}\right)t}$

Step 4. we get:

$1=\begin{array}{l}\frac{B^2{\ell }^2}{R.2m}t\end{array}$

$\begin{array}{l}t=\frac{2mR}{B^2{\ell }^2}\\ t=\frac{2\times 50\times 10}{{\left(0.1\right)}^2\times {10}^2}\end{array}$

$t=10000s$

Hence, the correct option is (B).