Find the value of magnetic field between plates of capacitor at a distance 1 ~m from centre, where electric field varies by 10^10 ~V / m per second.

Correct answer: (A) 5.56 × 10^8 ~T. Magnetic field B =fracμ_0 ε_0 r2 (d E)/(d t)=frac19 × 10^16 × 2 × 10^10 =5.56 × 10^-8 ~T

Q106 / 120 · VITEEE 2010 · Physics

Find the value of magnetic field between plates of capacitor at a distance $1 \mathrm{~m}$ from centre, where electric field varies by $10^{10} \mathrm{~V} / \mathrm{m}$ per second.

Correct answer
A
$5.56 \times 10^{8} \mathrm{~T}$
B
$5.56 \times 10^{-3} \mathrm{~T}$
C
$5.56 \mu \mathrm{T}$
D
$5.55 \mathrm{~T}$

Solution

Magnetic field $$ \begin{aligned} B &=\frac{\mu_{0} \varepsilon_{0} r}{2} \frac{d E}{d t}=\frac{1}{9 \times 10^{16} \times 2} \times 10^{10} \\ &=5.56 \times 10^{-8} \mathrm{~T} \end{aligned} $$

Hence, the correct answer is option: 'A'.

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