Solution:
The magnetic field inside a solenoid is given by \( B = \mu_0 \frac{N}{L} I \). Calculating the ratio: \( \frac{B_A}{B_B} = \frac{N_A / L_A}{N_B / L_B} = \frac{N / 2L}{2N / 3L} = \frac{3}{4} \).
The magnetic field inside a solenoid is given by \( B = \mu_0 \frac{N}{L} I \). Calculating the ratio: \( \frac{B_A}{B_B} = \frac{N_A / L_A}{N_B / L_B} = \frac{N / 2L}{2N / 3L} = \frac{3}{4} \).
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