There is a long cylinder of radius R having a cylindrical cavity of radius R/2 as shown in the figure. A part from the cavity, the entire space inside the cylinder has a uniform magnetic field parallel to the axis of the cylinder. The magnetic field starts changing at a uniform rate of $\frac{dB}{dt} = k T / s$ . If the induced electric field at a point inside cavity is kR/p. Then P is?

Question

There is a long cylinder of radius R having a cylindrical cavity of radius R/2 as shown in the figure. A part from the cavity, the entire space inside the cylinder has a uniform magnetic field parallel to the axis of the cylinder. The magnetic field starts changing at a uniform rate of $\frac{dB}{dt} = k T / s$ . If the induced electric field at a point inside cavity is kR/p. Then P is?

Infinity Learn · Accepted Answer

We can consider the entire system of magnetic field to be superposition of fields as shows below. $\frac{{dB}_{1}}{dt} = \frac{{dB}_{2}}{dt} = α$

Say, both fields are increasing

Induced electric field inside a cylindrical region having time changing magnetic field is given by $E = \frac{r}{2} (\frac{dB}{dt})$ where r is distance from the axis of the cylinder. Induced field at point P(inside cavity)
i) Due to changing $\vec{B_{1}}$ is $\vec{E_{1}}$ (perpendicular to $O_{1} P$ ) ${\vec{E}}_{1} = \frac{r_{1}}{2} \frac{{dB}_{1}}{dt} = \frac{r_{1}}{2} k$
ii) Due to changing ${\vec{B}}_{2} is {\vec{E}}_{2}$ (perpendicular to $O_{1} P) {\vec{E}}_{2} = \frac{r_{2}}{2} \frac{{dB}_{2}}{dt} = \frac{r_{2}}{2} k$

$E_{1} sinα = \frac{r_{1}}{2} ksinα$

$E_{2} sinβ = \frac{r_{2}}{2} ksinβ$

$E_{1} cosα = \frac{r_{1}}{2} kcosα$

$E_{2} cosβ = \frac{r_{2}}{2} kcosβ$

From geometry, $r_{1} cosα + r_{2} cosβ = \frac{R}{2} and r_{1} sinα = r_{2} sinβ$

Thus, $E_{1} sinα = E_{2} sinβ$

and $E_{net} = E_{1} cosα + E_{2} cosβ = \frac{k}{2} (r_{1} cosα + r_{2} cosβ) = \frac{kR}{4}$

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