Applications involving charged particles moving in a magnetic field

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Question

A current carrying loop lies on a smooth horizontal plane.Then,

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Solution

As the loop is placed in horizontal plane, so area vector is along vertical direction. From $\vec{τ} = I (\vec{A} \times \vec{B})$ , as $\vec{A}$ is in vertical direction, i would be the plane of loop only. So, option (1) is wrong because for rotation of loop about its own axis $\vec{τ}$ must be along vertical direction. (2) is correct because we can produce torque in the plane of the loop and due to this the loop can tip over.

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Similar Questions

A non-conducting non-magnetic rod having circular cross section of radius R is suspended from a rigid support as shown in the figure. A light and small coil of 300 turns is wrapped tightly at the left end of the rod where uniform magnetic field B exists in vertically downward direction.Air of density $ρ$ hits the half of the right part of the rod with velocity V as shown in the figure. What should be current in clockwise direction (as seen from O) in the coil so that rod remains horizontal? Give answer in m A. Given

$\frac{2}{L v} \sqrt{\frac{π R B}{ρ}} = \frac{1}{\sqrt{5}} A^{- 1 / 2}$