The mutual inductance between a primary and secondary circuits is 0.5 H. The resistance of the
primary and the secondary circuits are 20Ω and 5Ω respectively. To generate a current of 0.4 A
in the secondary, current in the primary must be changed at the rate of :
Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross-sectional area A = 10 cm² and length = 20 cm. If one of the solenoids has 300 turns and the other 400 turns, their mutual inductance is
\[\left( \mu=4\pi\times 10^{-7} T m A^{-1}\right)\] :
What is the mutual inductance of a two-loop system as shown with centre separation l ?
Two circular coils can be arranged in any of the three situations shown in the figure. Their
mutual inductance will be :
Based on the visual arrangement of the coils, the mutual inductance is maximum in situation (a).
This is because the coils are placed co-axially (one above the other), allowing the maximum amount of magnetic flux from the primary coil to pass through the secondary coil, resulting in the highest coupling coefficient.
Two coils X and Y are placed in a circuit such that a current changes by 2 A in coil X and the magnetic flux change of 0.4 Wb occurs in coil Y. The value of mutual inductance of coils is :
The mutual inductance ($M$) is calculated by the ratio of flux change in coil Y to the current change in coil X:
$M = 0.2 \text{ H}$
A small square loop of wire of side l is placed inside a large square loop of wire of side L ( L > l ). The loops are coplanar and their centres coincide. The mutual inductance of the system is proportional to :
The mutual inductance M is found by calculating the magnetic flux through the small loop due to the current I in the large loop. Using the formula for the magnetic field at the center of a square loop, B \frac{I}{L}.
Since the small loop is much smaller than the large one \(L \gg l)\, the field is approximately uniform across its area \A = l^2\. The flux \Phi = B \cdot A$ is therefore proportional to \\frac{I}{L} \cdot l^2\.
Because $M = \frac{\Phi}{I}$, the mutual inductance scales as:
Two coils of self inductance \(2\text{ mH}\) and \(8\text{ mH}\) are placed so close together that the effective flux in one coil is completely linked with the other. The mutual inductance between these coils is:
For complete coupling, the coupling coefficient \(k = 1\). The mutual inductance is given by \(M = k\sqrt{L_1 L_2} = \sqrt{2 \times 8} = 4\text{ mH}\).
A square coil of side length \(l\) is placed at centre of a large circular coil of radius \(R\), where \(R \gg l\) and coils are in same plane. The coefficient of mutual inductance of the coils is
Let a current \(I\) flow through the large circular coil, producing a magnetic field \(B = \frac{mu_0 I}{2R}\) at its center. The flux through the small square coil is \(\Phi = B'A = \left(\frac{\mu_0 I}{2R}\right) l^2\). Therefore, the mutual inductance is \(M = \frac{\Phi}{I} = \frac{\mu_0 l^2}{2R}\).
Two conducting circular loops of radii \(R_1\) and \(R_2\) are placed in the same plane with their centres coinciding. If \(R_1 \gg R_2\), the mutual inductance \(M\) between them will be directly proportional to
The magnetic field produced by the larger loop 1 at the center is \(B_1 = \frac{\mu_0 I_1}{2 R_1}\). The magnetic flux through the smaller loop 2 is \(\phi_2 = B_1 A_2 = \frac{\mu_0 I_1}{2 R_1} \pi R_2^2\). Therefore, \(M = \frac{\phi_2}{I_1} = \frac{\mu_0 \pi R_2^2}{2 R_1}\), which means \(M \propto \frac{R_2^2}{R_1}\).
Assertion (A): An emf is induced in a closed loop where magnetic flux is varied. The induced electric field is not a conservative field.
Reason (R): For induced electric field, the line integral \( \oint \vec{E} \cdot d\vec{l} \) around a closed path is non-zero.
Assertion (A) is true because varying magnetic flux induces a non-conservative electric field. Reason (R) is true; for a non-conservative field, \( \oint \vec{E} \cdot d\vec{l} \neq 0 \). (R) correctly explains why the induced electric field is non-conservative.