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JEE Main 2025 (22 Jan, Shift 2) · Moving Charge in a Magnetic Field

Moving Charge in a Magnetic Field question from JEE Main 2025 (22 Jan, Shift 2)

Q22JEE Main 2025 · 22 JAN · SHIFT 2Medium

Moving Charge in a Magnetic Field

A proton is moving undeflected in a region of crossed electric and magnetic fields at a constant speed of 2×105 ms−12 \times 10^{5}\text{ ms}^{-1}. When the electric field is switched off, the proton moves along a circular path of radius 2 cm. The magnitude of electric field is x×104 N/Cx \times 10^{4}\text{ N/C}. The value of xx is _________. Take the mass of the proton =1.6×10−27 kg= 1.6 \times 10^{-27}\text{ kg}.

Numerical answer type. Work it out, then open the answer.

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Answer: 2

  1. Circular path after EE is off: r=mvqB⇒B=mvqrr = \dfrac{mv}{qB} \Rightarrow B = \dfrac{mv}{qr}.
  2. B=1.6×10−27×2×1051.6×10−19×0.02=0.1 TB = \dfrac{1.6 \times 10^{-27} \times 2 \times 10^{5}}{1.6 \times 10^{-19} \times 0.02} = 0.1\text{ T}
  3. Undeflected motion: E=vB=2×105×0.1=2×104 N/CE = vB = 2 \times 10^{5} \times 0.1 = 2 \times 10^{4}\text{ N/C}
  4. So x=2x = 2.
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