A loop of flexible conducting wire lies in a magnetic field of $2.0 \mathrm{~T}$ with its plane perpendicular to the field. The length of the wire is $1 \mathrm{~m}$. When a current of $1.1 \mathrm{~A}$ is passed through the loop, it opens into a circle, then the tension developed in the wire is

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Aicharya · Accepted Answer

Let the tension developed is $\mathrm{T}$. Let radius of loop be $r$. For smaller element

$\begin{aligned} & \mathrm{BI}(\mathrm{d} l)=2 \mathrm{~T} \sin \left(\frac{d \theta}{2}\right) \\ & \Rightarrow \quad \mathrm{BI}(r \mathrm{~d} \theta)=2 \mathrm{~T}\left(\frac{d \theta}{2}\right) \end{aligned}$ $\theta$ is small $\sin \simeq \theta$ $\therefore \quad \mathrm{T}=i r \mathrm{~B}=\frac{i l B}{2 \pi}=\frac{1.1 \times 1 \times 2.0}{2 \times 3.14}=0.35 \mathrm{~N}$

A loop of flexible conducting wire lies in a magnetic field of $2.0 \mathrm{~T}$ with its plane…

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