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 ampere current passes through the rod it rises by $1\text{cm}$. Determine the magnetic field assuming acceleration due to gravity be $10{\text{m/s}}^2$.

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

Answer 1

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

Initially the rod will be in equilibrium if
$2k{x}_0=Mg$ then $x_0=4\text{cm}$
Now when the current $I$ is passed through the rod it will experience a force
$F=BIL$ vertically up; so in this situation for its equilibrium,
$2kx+BIL=Mg$
$(x=4-1=3\text{cm})$
So
$\frac{x}{x_0}=\frac{Mg-BIL}{Mg}$
$\Rightarrow \frac{x}{x_0}=1-\frac{BIL}{Mg}$
$\Rightarrow B=\frac{Mg(x_0-x)}{IL{x}_0}=\frac{10\times {10}^{-3}\times 10\times 1\times {10}^{-2}}{20\times 25\times {10}^{-2}\times 4\times {10}^{-2}}$
$=0.5\times {10}^{-2}\text{T}$