The number of turns of long solenoid is 500 the area of cross section of solenoid is 2 * 10^-3 m^2 if the value of magnetic induction
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Magnetic flux of toroid having n number of turns is
= nBA
B -> magnetic induction
A-> area of cross section
= 500 × 2 × 10^(-3) × 5 × 10^(-3)
= 10³ × 10^(-3) × 5 × 10^(-3)
= 5 × 10^(-3)
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Answer:
5*10^-3
Explanation:
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