Answer:
vi) Double the current in the wire, and double the number of turns in the 20-cm long solenoid
Explanation:
The magnetic field inside the solenoid and the current flowing in the coil of solenoid are related to each other by the following equation
B₀=μ₀nI₀
Where,
B₀ is the magnetic field in the middle of solenoid
n is the number of turns in the coil of solenoid
I₀ is the current flowing in the coil of solenoid
In the above equation, as μ₀ is a constant so the magnetic field will be directly proportional to the number of turns multiplied by the current. So, changing the radius of the coil or length of the coil will have no effect on the magnetic field.
As we have to increase the magnetic field by 4 times, we need to double the current as well as the number of turns as mentioned in the option vi.
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Idk if this is related to what you ask but it might help.
Answer:
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Answer:
32.76 Volt
Explanation:
frequency, f = 400 Hz
Area of crossection, A = 13 cm²
Maximum flux density, B = 0.9 tesla
Number of turns in secondary coil, N = 70
Let the maximum induced voltage is e.
According to the Faraday's law of electromagnetic induction, the induced emf is equal to the rate of change of magnetic flux.
e = dФ/dt

Time is defined as the reciprocal of frequency.
So, e = N B A f
e = 70 x 0.9 x 13 x 10^-4 x 400
e = 32.76 volt