Two conducting circular loops of radii R_1 an | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The magnetic field $B$ at the center of a circular loop of radius $R_1$ carrying current $I_1$ is given by $B = \frac{\mu_0 I_1}{2 R_1}$.Since $R_

Vedclass · Accepted Answer

$\frac{R_2^2}{R_1}$

Vedclass · Answer

(D) The magnetic field $B$ at the center of a circular loop of radius $R_1$ carrying current $I_1$ is given by $B = \frac{\mu_0 I_1}{2 R_1}$.Since $R_1$  >  $R_2$, we assume the smaller loop of radius $R_2$ is placed within the magnetic field of the larger loop.The magnetic flux $\phi_2$ linked with the smaller loop is $\phi_2 = B \cdot A_2$, where $A_2 = \pi R_2^2$ is the area of the smaller loop.Substituting the values, we get $\phi_2 = \left( \frac{\mu_0 I_1}{2 R_1} \right) (\pi R_2^2) = \left( \frac{\mu_0 \pi R_2^2}{2 R_1} \right) I_1$.By definition, the mutual inductance $M$ is given by $\phi_2 = M I_1$, so $M = \frac{\mu_0 \pi R_2^2}{2 R_1}$.Thus, $M$ is directly proportional to $\frac{R_2^2}{R_1}$.

Two conducting circular loops of radii $R_1$ and $R_2$ are placed in the same plane with their centres coinciding. If $R_1$ > $R_2$, the mutual inductance $M$ between them will be directly proportional to

Similar Questions

Which is the best method to conserve the wild relatives of plants?

Which one of the following is most effective in causing the coagulation of an $As_2S_3$ sol?

If $\frac{x^3}{(2x-1)(x+2)(x-3)} = A + \frac{B}{2x-1} + \frac{C}{x+2} + \frac{D}{x-3}$, then $A$ is equal to

$A$ body is executing simple harmonic motion. At a displacement $x$ its potential energy is $E_1$ and at a displacement $y$ its potential energy is $E_2$. The potential energy $(E)$ at displacement $(x+y)$ is

If the lines $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and $\frac{x - 3}{1} = \frac{y - k}{2} = \frac{z}{1}$ intersect,then $k =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module