Question

Figure shows a circular region of radius R in which
uniform magnetic field B exists. The magnetic field
dB
is increasing at a rate . The induced electric
field at a distance from the centre for r < R is ​

Answer 1

Answer:

$\bullet\;\;\boxed{\sf E=\dfrac{r}{2}\;.\;\bigg(\dfrac{dB}{dt}\bigg)}$

Explanation:

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As given the Magnetic field is increasing at a rate of dB/dt, therefore the magnetic field is not constant. Ans we need to find electric field inside the circular region as r < R. So, from the relation we know that,

$\longrightarrow\sf V=\displaystyle\oint\sf E\;.\;ds$

Here,

• V Denotes Voltage/EMF.
• E Denotes electric field.
• ds denotes small elemental length/distance.

Solving further,

$\longrightarrow\sf \dfrac{d \phi}{dt}=\displaystyle\oint\sf E\;.\;ds\ \ \ \ \ \because\Bigg[\dfrac{d\phi}{dt}=V\Bigg]\\\\\\\\\longrightarrow\sf \dfrac{d \phi}{dt}=E\displaystyle\oint\sf ds\\\\\\\\\longrightarrow\sf \dfrac{d\big(B\;.\;A\big)}{dt}=E\displaystyle\oint\sf ds\\\\\\\\\longrightarrow\sf E\displaystyle\oint\sf ds=A\;.\;\dfrac{dB}{dt}\\\\\\\\\longrightarrow\sf E\displaystyle\oint\sf ds=\pi r^{2}\;.\;\dfrac{dB}{dt}$

• ∫ ds = 2πr as it will cover whole length of the circle i.e. circumference.

$\longrightarrow\sf E\times 2\pi r=\pi r^{2}\;.\;\dfrac{dB}{dt}\\\\\\\\\longrightarrow\sf E=\dfrac{\pi r^{2}}{2\pi r}\;.\;\Bigg(\dfrac{dB}{dt}\Bigg)\\\\\\\\\longrightarrow\sf E=\dfrac{r}{2}\;.\;\Bigg(\dfrac{dB}{dt}\Bigg)\\\\\\\\\longrightarrow\large{\underline{\boxed{\red{\sf E=\dfrac{r}{2}\;.\;\Bigg(\dfrac{dB}{dt}\Bigg)}}}}$

Hence, we got the value of electric field.

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