If the length of a current-carrying solenoid is doubled keeping the number of turns same and the current is increased to 9 times, the value of magnetic field compared to the original magnetic field B will be:
Correct answer: B. 9/2 B
- A. 2/9 B
- B. 9/2 B
- C. 18 B
- D. 8/18
Explanation
The magnetic field inside a solenoid is given by:B = μ₀nIWhere:B is the magnetic field strengthμ₀ is the permeability of free spacen is the number of turns per unit lengthI is the current flowing through the solenoidIf the length of the solenoid is doubled, the number of turns per unit length (n) will be halved. If the current is increased to 9 times, the magnetic field strength will be:B' = μ₀(n/2)(9I) = (9/2)μ₀nI = (9/2)BTherefore, the magnetic field strength will be 9/2 times the original magnetic field strength.
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