Question

A current carrying conductor of mass $50\text{gm}$, length $0.5\text{m}$ carrying a current of $1\text{A}$ hangs by two identical springs each of spring constant $50\text{N/m}$ as shown in the figure below. If an outwards uniform magnetic field of $1\text{T}$ is applied on the conductor then the deformation in each spring will be ($g=10{\text{ms}}^{-2})$

• A. $2.5\text{cm}$
• B. $1.5\text{cm}$
• C. $1.0\text{cm}$
• D. $0.5\text{cm}$

Answer 1

The correct option is C $1.0\text{cm}$


The force on the current carrying conductor is in downward direction

$\therefore F_{spring}=\text{Weight}+F_{mag}$

$Kx+Kx=mg+Bil$

$\Rightarrow x=\frac{mg+Bil}{2K}$

Substituting the given data we get,

$x=\frac{(50\times {10}^{-3}\times 10)+(1\times 1\times 0.5)}{2\times 50}=\frac{0.5+0.5}{100}$

$=0.01\text{m}\text{or}1\text{cm}$

Hence, option (c) is the correct answer.