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A short solenoid of radius a, number of ...

A short solenoid of radius a, number of turns per unit length `n_(1)`, and length L is kept coaxially inside a very long solenoid of radius b, number of turns per unit lengt `n_(2)`. What is the mutual inductance of the system

A

`mu_(0)pib^(2)n_(1)n_(2)L`

B

`mu_(0)pia^(2)n_(1)n_(2)L^(2)`

C

`mu_(0)pia^(2)n_(1)n_(2)L`

D

`mu_(0)pib^(2)n_(1)n_(2)L^(2)`

Text Solution

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The correct Answer is:
To find the mutual inductance \( M \) of a short solenoid placed inside a long solenoid, we can follow these steps: ### Step 1: Understand the Configuration We have two solenoids: - A short solenoid of radius \( a \), number of turns per unit length \( n_1 \), and length \( L \). - A long solenoid of radius \( b \), number of turns per unit length \( n_2 \). ### Step 2: Area of the Short Solenoid The area \( A \) of the cross-section of the short solenoid is given by: \[ A = \pi a^2 \] ### Step 3: Magnetic Field Inside the Long Solenoid The magnetic field \( B \) inside the long solenoid (which is uniform) is given by: \[ B = \mu_0 n_2 I_2 \] where \( I_2 \) is the current flowing through the long solenoid, and \( \mu_0 \) is the permeability of free space. ### Step 4: Magnetic Flux Through the Short Solenoid The magnetic flux \( \Phi \) through the short solenoid due to the magnetic field of the long solenoid is: \[ \Phi = B \cdot A = (\mu_0 n_2 I_2) \cdot (\pi a^2) = \mu_0 \pi a^2 n_2 I_2 \] ### Step 5: Induced EMF and Mutual Inductance The mutual inductance \( M \) can be defined as the ratio of the magnetic flux linkage \( \Phi \) to the current \( I_2 \): \[ M = \frac{\Phi}{I_2} = \frac{\mu_0 \pi a^2 n_2 I_2}{I_2} = \mu_0 \pi a^2 n_2 \] ### Step 6: Total Mutual Inductance Considering Short Solenoid Since the short solenoid has \( n_1 \) turns per unit length, we need to consider the effect of \( n_1 \) on the mutual inductance: \[ M = \mu_0 \pi a^2 n_1 n_2 L \] ### Final Expression for Mutual Inductance Thus, the mutual inductance of the system is given by: \[ M = \mu_0 \pi a^2 n_1 n_2 L \] ### Conclusion The mutual inductance of the system is: \[ M = \mu_0 \pi a^2 n_1 n_2 L \]

To find the mutual inductance \( M \) of a short solenoid placed inside a long solenoid, we can follow these steps: ### Step 1: Understand the Configuration We have two solenoids: - A short solenoid of radius \( a \), number of turns per unit length \( n_1 \), and length \( L \). - A long solenoid of radius \( b \), number of turns per unit length \( n_2 \). ### Step 2: Area of the Short Solenoid ...
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