Induced Electric field

Electricity and Magnetism Level 5

The figure above shows a current I I flowing through a toroid with N = 1 , 000 N=1,000 turns. There will be a magnetic field in the toroid whose expression can be found through Ampere's loop law. The current changes according to the law d i d t = 5 \frac {di}{dt} = 5 amps. Find E E , the magnitude of the induced electric field at the centre of the figure.

Details and Assumptions

  • Approximate the toroid as having a fairly uniform radius which is to say assume r = 10 cm r=10 \text{ cm}
  • The radius of the inner portion of toroid is R = 1 cm R=1\text{cm}
  • μ 0 \mu _{ 0 } = 4 π × 1 0 − 7 \displaystyle 4 \pi \times 10^{-7}


The answer is 0.005.

Mvs Saketh by Mvs Saketh , 24, India

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1 solution

Jatin Yadav
Dec 15, 2015

In my approach, I consider N N rings on the toroid and find the electric fields due to them, then sum them. The contribution from each of the N N rings at the center of the Torus are equivalent.

Thus, we have d E × 2 π r = π R 2 μ 0 d I d t N 2 π R d n dE \times 2 \pi r = \dfrac{\pi R^2 \mu_{0} \frac{dI}{dt} N}{2 \pi R} dn .

Hence, we arrive at E = μ 0 N R 2 4 π r 2 ∫ d n \displaystyle E = \dfrac{\mu_{0} N R^2}{4 \pi r^2} \int dn and find E = μ 0 N 2 R 2 4 π r 2 d i d t E = \dfrac{\mu_{0} N^2 R^2}{4 \pi r^2} \dfrac{di}{dt}

3 Helpful 0 Interesting 0 Brilliant 1 Confused

Did you get E = 0.005 (where R R and r r follow the definition of question.)?

Lu Chee Ket - 5 years, 6 months ago

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