Q155 / 160 · AP EAMCET 2020 (21 Sep Shift 2) · Physics

A proton enters a magnetic field of flux density \(1.5 \mathrm{~Wb} \mathrm{~m}^{-2}\) with a velocity of \(2 \times 10^7 \mathrm{~ms}^{-1}\) at an angle of \(30^{\circ}\) with the field. The force on the proton will be

  • Correct answer
    A
    \(2.4 \times 10^{-12} \mathrm{~N}\)
  • B
    \(24 \times 10^{-12} \mathrm{~N}\)
  • C
    \(0.24 \times 10^{-12} \mathrm{~N}\)
  • D
    \(0.024 \times 10^{-12} \mathrm{~N}\)

Solution

Magnetic flux density, \(B=1.5 \mathrm{Wbm}^{-2}\) Velocity of proton, \(v=2 \times 10^7 \mathrm{~ms}^{-1}\) \(\theta=30^{\circ}\) Charge on proton, \(q=1.6 \times 10^{-19} \mathrm{C}\) \(\therefore\) Force on the proton, \(\begin{aligned} F & =B q v \sin \theta \\ & =1.5 \times 1.6 \times 10^{-19} \times 2 \times 10^7 \times \sin 30^{\circ} \\ & =4.8 \times 10^{-12} \times \frac{1}{2}=2.4 \times 10^{-12} \mathrm{~N} \end{aligned}\)

Hence, the correct answer is option: 'A'.

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