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JEE Advanced 2025 Paper 2 · Physics · Magnetic Effects of Current · hard

A proton (mass 1.67×10^-27 kg, charge 1.6×10^-19 C) is released from rest in a region with a uniform electric field E=2000 V m^-1 along +y and a…

Numerical answer

A proton (mass 1.67×10−27 kg1.67\times10^{-27}\ \mathrm{kg}, charge 1.6×10−19 C1.6\times10^{-19}\ \mathrm{C}) is released from rest in a region with a uniform electric field E=2000 V m−1E=2000\ \mathrm{V\,m^{-1}} along +y+y and a uniform magnetic field B=0.10 TB=0.10\ \mathrm{T} along +z+z. Gravity is negligible. The proton traces a cycloid-like path. Find the length of the path travelled during one full period of its motion (from one point of zero speed to the next), in millimetres. Give the answer correct to two decimal places.
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Answer: 16.70

In crossed fields the motion is a cycloid: a uniform circular motion with angular frequency ω=qB/m\omega=qB/m superimposed on a drift of speed vd=E/Bv_d=E/B along xx. Starting from rest, the position is x=a(ωt−sin⁡ωt)x=a(\omega t-\sin\omega t), y=a(1−cos⁡ωt)y=a(1-\cos\omega t) with a=vdω=mEqB2a=\dfrac{v_d}{\omega}=\dfrac{mE}{qB^2}. Numerically a=1.67×10−27×20001.6×10−19×0.01=2.0875×10−3 ma=\dfrac{1.67\times10^{-27}\times2000}{1.6\times10^{-19}\times0.01}=2.0875\times10^{-3}\ \mathrm{m}. The speed is ∣v∣=2vdsin⁡(ωt/2)|v|=2v_d\sin(\omega t/2), so the arc length of one arch is s=∫02π/ω2vdsin⁡(ωt/2) dt=8vdω=8as=\int_0^{2\pi/\omega}2v_d\sin(\omega t/2)\,dt=\dfrac{8v_d}{\omega}=8a. Therefore s=8×2.0875 mm=16.70 mms=8\times2.0875\ \mathrm{mm}=16.70\ \mathrm{mm}.
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