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JEE Advanced
Exam
JEE Advanced
Subject
Physics
Chapter
Mechanical Properties of Solids

JEE Advanced · Physics · Mechanical Properties of Solids

A coil carrying a current I=10 ~mA is placed in a uniform magnetic...

A coil carrying a current I=10 mAI=10 \mathrm{~mA} is placed in a uniform magnetic field so that its axis coincides with the field direction. The coil consist of only one turn and made up of copper. The diameter of the wire is 0.1 mm , the radius of the coil is R=3πcmR=3 \pi \mathrm{cm}. The value of external induction field will the coil rupture is 500 K . Breaking stress of copper =3×108 N/m2=3 \times 10^8 \mathrm{~N} / \mathrm{m}^2. Then find the value of KK.A coil carrying a current I=10 ~mA is placed in a uniform magnetic field so that its axis coincides with the field

Correct answer

5

Step-by-step solution

2Tsin⁡Δθ2=Bi(ΔθR)⇒T=IRB also, T=πr2σ∴B=πr2σIR\begin{aligned} & 2 T \sin \frac{\Delta \theta}{2}=\mathrm{Bi}(\Delta \theta \mathrm{R}) \\ & \Rightarrow \mathrm{T}=\mathrm{IRB} \\ & \text { also, } T=\pi \mathrm{r}^2 \sigma \\ & \therefore \mathrm{B}=\frac{\pi \mathrm{r}^2 \sigma}{\mathrm{IR}}\end{aligned} A coil carrying a current I=10 ~mA is placed in a uniform magnetic field so that its axis coincides with the field
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