Question

A current carrying wire of length $l$ is placed near an infinitely long, straight conductor, as shown. What will the work done to move this wire away from the conductor through $4\text{cm}$. Neglect the effect of gravity.
Given: $I_1=10\text{A},I=2\text{A},l=10\text{cm},\ln 3=1.1$

• A. $2.2\times {10}^{-7}\text{J}$
• B. $1.1\times {10}^{-7}\text{J}$
• C. $3.3\times {10}^{-7}\text{J}$
• D. $4.4\times {10}^{-7}\text{J}$

Answer 1

The correct option is D $4.4\times {10}^{-7}\text{J}$

As the wire moves away from the conductor, the magnitude of field will go on decreasing so that the force will also decrease.


Let the wire is moved away through a distance $dx$, then the work done on the wire will be

$dW=Fdx$

$W={\int }_{2cm}^{6cm}Fdx........\left(1\right)$

Here, the force $F$ will have the same magnitude as that of $F_m$.

Magnetic force at the wire for given situation will be

$F_m=ilB=Il(\frac{{\mu }_0{I}_1}{2\pi x})$

$\therefore F=\left(\frac{{\mu }_{0}I{I}_{1}l}{2\pi }\right)\frac{1}{x}$

Substituting this vale in equation $\left(1\right)$, we get

$W={\int }_{2cm}^{6cm}\left(\frac{{\mu }_{0}I{I}_{1}l}{2\pi }\right)\frac{dx}{x}$

$W=\left(\frac{{\mu }_{0}I{I}_{1}l}{2\pi }\right)\ln x{|}_2^6=\left(\frac{{\mu }_{0}I{I}_{1}l}{2\pi }\right)\ln 3$

$W=\left(\frac{4\pi \times {10}^{-7}\times 2\times 10\times (10\times {10}^{-2})}{2\pi }\right)\ln 3$

$\therefore W=4.4\times {10}^{-7}\text{J}$

Therefore, option $\left(D\right)$ is correct.

Alternate sol: Instead of using $F=Bil$, we can also use the expression of force per unit length between the two parallel wires, which will give the same result.