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Magnetics question

2023 · 1 Feb · Shift 1 · Q64
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Magnetics question

2023 · 1 Feb · Shift 1 · Q64

JEE MainPhysicsMagneticsNumerical+4 / −1
A charge particle of 2 μC2 ~\mu \mathrm{C}2 μC accelerated by a potential difference of 100 V100 \mathrm{~V}100 V enters a region of uniform magnetic field of magnitude 4 mT4 ~\mathrm{mT}4 mT at right angle to the direction of field. The charge particle completes semicircle of radius 3 cm3 \mathrm{~cm}3 cm inside magnetic field. The mass of the charge particle is ‾\underline{\hspace{2cm}}​×10−18 kg\times 10^{-18} \mathrm{~kg}×10−18 kg
Numerical answer
View written solutionFree

Correct answer: 144

  1. Given data
  • Charge: q=2 μC=2×10−6 Cq = 2\,\mu\text{C} = 2\times 10^{-6}\,\text{C}q=2μC=2×10−6C
  • Accelerating potential: V=100 VV = 100\,\text{V}V=100V
  • Magnetic field: B=4 mT=4×10−3 TB = 4\,\text{mT} = 4\times 10^{-3}\,\text{T}B=4mT=4×10−3T
  • Radius of semicircle: r=3 cm=3×10−2 mr = 3\,\text{cm} = 3\times 10^{-2}\,\text{m}r=3cm=3×10−2m

We need to find the mass mmm in the form: m=‾×10−18 kgm = \underline{\hspace{1cm}} \times 10^{-18}\,\text{kg}m=​×10−18kg

  1. Speed gained by acceleration through potential difference

When a charged particle is accelerated through potential difference VVV, the gained kinetic energy is: qV=12mv2qV = \frac{1}{2}mv^2qV=21​mv2

So, v=2qVmv = \sqrt{\frac{2qV}{m}}v=m2qV​​

  1. Motion in magnetic field

Since the particle enters perpendicular to the magnetic field, magnetic force provides centripetal force: qvB=mv2rqvB = \frac{mv^2}{r}qvB=rmv2​

Cancel one vvv: qB=mvrqB = \frac{mv}{r}qB=rmv​ v=qBrmv = \frac{qBr}{m}v=mqBr​

  1. Substitute into energy equation

From step 2: qV=12mv2qV = \frac{1}{2}mv^2qV=21​mv2

Using v=qBrmv = \frac{qBr}{m}v=mqBr​

we get qV=12m(qBrm)2qV = \frac{1}{2}m\left(\frac{qBr}{m}\right)^2qV=21​m(mqBr​)2

qV=12m⋅q2B2r2m2qV = \frac{1}{2}m\cdot \frac{q^2B^2r^2}{m^2}qV=21​m⋅m2q2B2r2​

qV=q2B2r22mqV = \frac{q^2B^2r^2}{2m}qV=2mq2B2r2​

Therefore, m=qB2r22Vm = \frac{qB^2r^2}{2V}m=2VqB2r2​

  1. Put the values

m=(2×10−6)(4×10−3)2(3×10−2)22×100m = \frac{(2\times 10^{-6})(4\times 10^{-3})^2(3\times 10^{-2})^2}{2\times 100}m=2×100(2×10−6)(4×10−3)2(3×10−2)2​

Now simplify step by step:

(4×10−3)2=16×10−6(4\times 10^{-3})^2 = 16\times 10^{-6}(4×10−3)2=16×10−6 (3×10−2)2=9×10−4(3\times 10^{-2})^2 = 9\times 10^{-4}(3×10−2)2=9×10−4

So, m=(2×10−6)(16×10−6)(9×10−4)200m = \frac{(2\times 10^{-6})(16\times 10^{-6})(9\times 10^{-4})}{200}m=200(2×10−6)(16×10−6)(9×10−4)​

Multiply coefficients: 2×16×9=2882\times 16\times 9 = 2882×16×9=288

Multiply powers of 10: 10−6⋅10−6⋅10−4=10−1610^{-6}\cdot 10^{-6}\cdot 10^{-4} = 10^{-16}10−6⋅10−6⋅10−4=10−16

Thus, m=288×10−16200m = \frac{288\times 10^{-16}}{200}m=200288×10−16​

m=1.44×10−16 kgm = 1.44\times 10^{-16}\,\text{kg}m=1.44×10−16kg

Now write in the required form: m=144×10−18 kgm = 144\times 10^{-18}\,\text{kg}m=144×10−18kg

  1. Final answer

The required integer is: 144\boxed{144}144​

  1. Comparison with stored answer

Stored correct answer = 144144144

Our derived answer = 144144144

Hence, the answer agrees with the stored correct answer.

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