Magnetic Effects & Magnetism

JEE Main level, single correct, medium

78 of 276

Question 78 of 276

A thin ring of 10cm10cm radius carries a uniformly distributed charge. The ring rotates at a constant angular speed of 40πrads−140\pi rad{s}^{-1} about its axis, perpendicular to its plane. Is the magnetic field its centre is 3.8×10−9T3.8\times {10}^{-9}T , then the charge carried by the ring is close to (μ0=4π×10−7N/A2).\left({\mu }_{0}=4\pi \times {10}^{-7}N/{A}^{2}\right).

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Answer: (B) 3×10−5C3\times {10}^{-5}C

i=qt=q2π/ω=qω2πi=\dfrac{q}{t}=\dfrac{q}{2\pi /\omega }=\dfrac{q\omega }{2\pi }
B=μ0i2R=μ0qω4πRB=\dfrac{{\mu }_{0}i}{2R}=\dfrac{{\mu }_{0}q\omega }{4\pi R}
4π×10−7×q×40π4π×(0.1)=3.8×10−9\dfrac{4\pi \times {10}^{-7}\times q\times 40\pi }{4\pi \times (0.1)}=3.8\times {10}^{-9}
q×(400π)=3.8×10−2q\times \left(400\pi \right)=3.8\times {10}^{-2}
q=3×10−5q=3\times {10}^{-5} C

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