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Roman55 [17]
3 years ago
9

In physics grade 10 number 8 in the pic how do we do it thanks in advance

Physics
1 answer:
Rom4ik [11]3 years ago
5 0
Hope this helps you.

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Read 2 more answers
When the current in a toroidal solenoid is changing at a rate of 0.0240 A/s , the magnitude of the induced emf is 12.4 mV . When
Gemiola [76]

Answer:

The number of turns in the solenoid is 230.

Explanation:

Given that,

Rate of change of current, \dfrac{dI}{dt}=0.0240\ A/s

Induced emf, \epsilon=12.4\ mV=12.4\times 10^{-3}\ V

Current, I = 1.5 A

Magnetic flux, \phi=0.00338\ Wb

The induced emf through the solenoid is given by :

\epsilon=L\dfrac{dI}{dt}

or

L=\dfrac{\epsilon}{(di/dt)}........(1)

The self inductance of the solenoid is given by :

L=\dfrac{N\phi}{I}.........(2)

From equation (1) and (2) we get :

\dfrac{\epsilon}{(di/dt)}=\dfrac{N\phi}{I}

N is the number of turns in the solenoid

N=\dfrac{\epsilon I}{\phi (dI/dt)}

N=\dfrac{12.4\times 10^{-3}\times 1.5}{0.00338 \times 0.024}

N = 229.28 turns

or

N = 230 turns

So, the number of turns in the solenoid is 230. Hence, this is the required solution.

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3 years ago
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