A uniform magnetic field of induction B is confined in a cylindrical region of the radius R . If the field is increasing at a constant rate of (d B/d t)=α Ts- 1 , then the intensity of the electric field induced at a point P distant r from the axis as shown in the figure is equal to

Question

A uniform magnetic field of induction B is confined in a cylindrical region of the radius R . If the field is increasing at a constant rate of (d B/d t)=α Ts- 1 , then the intensity of the electric field induced at a point P distant r from the axis as shown in the figure is equal to <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-9zmmnya61qrtfcft.jpg" /> (A) (1/8)r (B) (1/8)rα  (C) (1/2)rα  (D) r

Tardigrade Admin · Accepted Answer

Magnetic flux linked with circular section of radius

r

of the cylinder ,

ϕ = (π r^{2}) B .

Due to this induced emf, an electric field E is established which is equal to potential difference per unit length.

E = \frac{e}{2 π r} = \frac{π r ^{2} α}{2 π r}

E = \frac{1}{2} r α

$\frac{1}{8} r$

$\frac{1}{8} r α$

$\frac{1}{2} r α$

$r$

Magnetic flux linked with circular section of radius $r$ of the cylinder , $ϕ = (π r^{2}) B .$

Due to this induced emf, an electric field E is established which is equal to potential difference per unit length.
$E = \frac{e}{2 π r} = \frac{π r ^{2} α}{2 π r}$
$E = \frac{1}{2} r α$

Q. A uniform magnetic field of induction B is confined in a cylindrical region of the radius R . If the field is increasing at a constant rate of dtdB​=αTs−1 , then the intensity of the electric field induced at a point P distant r from the axis as shown in the figure is equal to

81​r

81​rα

21​rα

r

Magnetic flux linked with circular section of radius r of the cylinder , ϕ=(πr2)B. Due to this induced emf, an electric field E is established which is equal to potential difference per unit length. E=2πre​=2πrπr2α​ E=21​rα

$\frac{1}{8} r$

$\frac{1}{8} r α$

$\frac{1}{2} r α$

$r$

Magnetic flux linked with circular section of radius $r$ of the cylinder , $ϕ = (π r^{2}) B .$

Due to this induced emf, an electric field E is established which is equal to potential difference per unit length.
$E = \frac{e}{2 π r} = \frac{π r ^{2} α}{2 π r}$
$E = \frac{1}{2} r α$