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Magnetics question

2019 · 12 Apr · Shift 1 · Q64
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Magnetics question

2019 · 12 Apr · Shift 1 · Q64

JEE MainPhysicsMagneticsMCQ+4 / −1
A thin ring of 10 cm radius carries a uniformly distributed charge. The ring rotates at a constant angular speed of 40 π\piπ rad s–1 about its axis, perpendicular to its plane. If the magnetic field at its centre is 3.8 × 10–9 T, then the charge carried by the ring is close to (μ\muμ 0 = 4 π\piπ × 10–7 N/A2 ).
  1. A
    7 × 10–6 C
  2. B
    4 × 10–5 C
  3. C
    2 × 10–6 C
  4. D
    3 × 10–5 C
View written solutionFree

Correct answer: D

  1. Magnetic field at the centre of a current-carrying ring

For a circular loop of radius RRR carrying current III, the magnetic field at the centre is

B=μ0I2RB = \frac{\mu_0 I}{2R}B=2Rμ0​I​
  1. Current due to rotating charged ring

If total charge QQQ is uniformly distributed on the ring and it rotates with angular speed ω\omegaω, then time period is

T=2πωT = \frac{2\pi}{\omega}T=ω2π​

Hence current,

I=QT=Qω2πI = \frac{Q}{T} = \frac{Q\omega}{2\pi}I=TQ​=2πQω​
  1. Substitute into magnetic field formula
B=μ02R⋅Qω2πB = \frac{\mu_0}{2R} \cdot \frac{Q\omega}{2\pi}B=2Rμ0​​⋅2πQω​

So,

B=μ0Qω4πRB = \frac{\mu_0 Q \omega}{4\pi R}B=4πRμ0​Qω​

Thus,

Q=4πRBμ0ωQ = \frac{4\pi R B}{\mu_0 \omega}Q=μ0​ω4πRB​
  1. Insert given values

Given:

R=10 cm=0.1 mR = 10\text{ cm} = 0.1\text{ m}R=10 cm=0.1 m ω=40π rad s−1\omega = 40\pi\text{ rad s}^{-1}ω=40π rad s−1 B=3.8×10−9 TB = 3.8 \times 10^{-9}\text{ T}B=3.8×10−9 T μ0=4π×10−7\mu_0 = 4\pi \times 10^{-7}μ0​=4π×10−7

Therefore,

Q=4π⋅0.1⋅3.8×10−9(4π×10−7)(40π)Q = \frac{4\pi \cdot 0.1 \cdot 3.8\times 10^{-9}}{(4\pi\times 10^{-7})(40\pi)}Q=(4π×10−7)(40π)4π⋅0.1⋅3.8×10−9​

Cancel 4π4\pi4π:

Q=0.1⋅3.8×10−9(10−7)(40π)Q = \frac{0.1\cdot 3.8\times 10^{-9}}{(10^{-7})(40\pi)}Q=(10−7)(40π)0.1⋅3.8×10−9​ Q=3.8×10−1040π×10−7Q = \frac{3.8\times 10^{-10}}{40\pi\times 10^{-7}}Q=40π×10−73.8×10−10​ Q=3.8×10−104π×10−6Q = \frac{3.8\times 10^{-10}}{4\pi\times 10^{-6}}Q=4π×10−63.8×10−10​ Q=3.84π×10−4Q = \frac{3.8}{4\pi}\times 10^{-4}Q=4π3.8​×10−4

Using π≈3.14\pi \approx 3.14π≈3.14,

Q≈3.812.56×10−4Q \approx \frac{3.8}{12.56}\times 10^{-4}Q≈12.563.8​×10−4 Q≈0.302×10−4=3.02×10−5 CQ \approx 0.302\times 10^{-4} = 3.02\times 10^{-5}\text{ C}Q≈0.302×10−4=3.02×10−5 C
  1. Closest option
Q≈3×10−5 CQ \approx 3 \times 10^{-5}\text{ C}Q≈3×10−5 C

So the correct option is D.

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