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QPages 249-253: Q46: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Due to a Current Pages 249-253: Q46 Solution: A charge of is distributed uniformly over a circular ring of…

A charge of 3.14×10−6 C3.14 \times 10^{-6} \text{ C} is distributed uniformly over a circular ring of radius 20.0 cm. The ring rotates about its axis with an angular velocity of 60.0 rad s−160.0 \text{ rad s}^{-1}. Find the ratio of the electric field to the magnetic field at a point on the axis at a distance of 5.00 cm from the centre.

Step-by-Step Verified Solution

**Answer:** E/B=1.8×109E/B = 1.8 \times 10^9 m/s. **Step-by-Step Explanation:** q=3.14×10−6q = 3.14 \times 10^{-6} C, r=0.2r = 0.2 m, ω=60\omega = 60 rad/s. i=qω2π=3×10−5i = \frac{q\omega}{2\pi} = 3 \times 10^{-5} A. B=μ0i2r=4π×10−7×3×10−52×0.2=9.4×10−11B = \frac{\mu_0 i}{2r} = \frac{4\pi \times 10^{-7} \times 3 \times 10^{-5}}{2 \times 0.2} = 9.4 \times 10^{-11} T. E=kqx(r2+x2)3/2=9×109×3.14×10−6×0.05(0.04+0.0025)3/2=1.7×105E = \frac{kqx}{(r^2 + x^2)^{3/2}} = \frac{9 \times 10^9 \times 3.14 \times 10^{-6} \times 0.05}{(0.04 + 0.0025)^{3/2}} = 1.7 \times 10^5 V/m. E/B=1.8×109E/B = 1.8 \times 10^9 m/s.

All Questions in this Chapter

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