Question

A metal rod of mass $10\text{gm}$ and length $25\text{cm}$ is suspended on two springs, as shown in figure. The springs are extended by $4\text{cm}$. When a $20\text{A}$ current passes through the rod, it rises by $1\text{cm}$. Determine the magnetic field , assuming acceleration due to gravity to be $10{\text{m/s}}^2$.

• A. $5\times {10}^{-5}\text{T}$
  
• B. $5\times {10}^{-4}\text{T}$
  
• C. $5\times {10}^{-2}\text{T}$
  
• D. $5\times {10}^{-3}\text{T}$
  

Answer 1

The correct option is D $5\times {10}^{-3}\text{T}$


Using $F_B=I(\overrightarrow{L}\times \overrightarrow{B})$, direction of magnetic force shown below in the figure.
​​​​
Initially when current, $I=0$, let the tension in spring be $T_0$ and extension be $x_0$. So,

$2T_0=mg....\left(1\right)$ and

$T_0=k{x}_0...\left(2\right)$

Given, $x_0=4\text{cm}$

Now when $20\text{A}$ current flows, the magnetic force acting on the rod,

$F_B=BIL\sin {90}^{\circ }$ (upwards)

Now, the new tension be,

$T=kx.....\left(3\right)$

As the rod rises by $1\text{cm}$

$\therefore x=x_0-1=4-1=3\text{cm}$

Now, from the free body diagram,

$2T+F_B=mg$

$\Rightarrow 2T=mg-BIL.....\left(4\right)$

Now, equation $\left(4\right)$ divided by $\left(1\right)$, we get

$\frac{T}{T_0}=\frac{mg-BIL}{mg}$

Using equations $\left(2\right)$ and $\left(3\right)$,

$\Rightarrow \frac{kx}{k{x}_0}=1-\frac{BIL}{mg}$

$\Rightarrow B=\frac{mg(x_0-x)}{IL{x}_0}$

Substituting the values,

$\Rightarrow B=\frac{10\times {10}^{-3}\times 10\times [4-3]\times {10}^{-2}}{20\times 25\times {10}^{{}^{-2}}\times 4\times {10}^{-2}}$

$\therefore B=5\times {10}^{-3}\text{T}$

Hence, option (d) is the correct answer.