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ExamsJEE AdvancedPhysics

A hypothetical radial magnetic field B = B0 * ur exists above the Earth's surface, where ur is a unit vector directed radially outward from a point on the Earth's surface (taken as the origin). A light charged particle undergoes uniform circular motion in a horizontal plane under the combined effect of Earth's gravity (g, downward) and this magnetic field. If the particle moves with speed v in a circle of radius r, the height h of the plane above the surface is h = n*v²/g. Find n.

  1. 1
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  3. 3
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Correct answer: 1

Solution

At any point on the circular orbit at height h and radius r, the position vector from origin has magnitude sqrt(r²+h²). The magnetic force F = qvB0/sqrt(r²+h²) * (h*r_hat_horizontal - r*z_hat). Vertical balance: qvB0*r/sqrt(r²+h²) = mg. Horizontal (centripetal): qvB0*h/sqrt(r²+h²) = mv²/r. Dividing: r/h = gr/v², so h = v²/g. Thus n = 1.

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