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The magnetic field due to a narrow solenoid 50 cm long with 4000 turns and current of 2a. |
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Answer» Magnetic field inside inside the solenoid \(B = \mu_0ni\) \(∵\) \(n = \frac Nl\) \(B = \frac{\mu_0Ni}{l}\) \(B = 4\pi \times 10^{-7} \times \frac{4000}{.50}\times 2\) \(B = 200960 \times 10^{-7}\) \(B = 2.0 \times 10^{-2} T\) Magnetic field at the end of solenoid \(B_{end} = \frac{\mu_0ni}{2}\) \(B_{end} = \frac{2.0 \times 10^{-2}}{2}\) \(B_{end} = 1 \times 10^{-2} T\) |
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