Question

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

• A. $0.5\times {10}^{-2}T$
• B. $5\times {10}^{-2}T$
• C. $5T$
• D. $10T$

Answer 1

The correct option is A $0.5\times {10}^{-2}T$

FB$=$BIL

initally$:$ weight$=$spring force

$\Rightarrow \frac{10}{1000}\times 10=2\times K\times \frac{4}{100}$

$\Rightarrow K=\frac{10}{8}N/m$

When current is passed

FB+spring force$=$ weight

$BIL+2\times \frac{10}{8}\times \frac{3}{100}=\frac{10}{100}\times 10$

$\Rightarrow B\times 20\times \frac{25}{100}=\frac{1}{10}-\frac{3}{40}$

$\Rightarrow 5B=\frac{1}{40}\Rightarrow B=\frac{1}{200}=0.5\times {10}^{-2}$

hence,

option $\left(A\right)$ is correct answer.