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Magnetic field at the centre of a long current carrying solenoid is $8.28 \times 10^{- 4} T$ . If length of the solenoid is halved and number of turns is doubled, what will be the magnetic field near the ends of the solenoid?

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4.14×10−4T

8.28×10−4T

16.56×10−4T

24.12×10−4T

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detailed solution

Correct option is C

At the centre B=μ0ni=μ0NliNear the ends, B1=μ02N1l1i∴B1B=12(N1N)(ll1)=12(2)(2)=2∴ B1=2×8.28×10−4T=16.56×10−4T

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Magnetic field at the centre of a long current carrying solenoid is 8.28×10−4T. If length of the solenoid is halved and number of turns is doubled, what will be the magnetic field near the ends of the solenoid?

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material

Magnetic field at the centre of a long current carrying solenoid is $8.28 \times 10^{- 4} T$ . If length of the solenoid is halved and number of turns is doubled, what will be the magnetic field near the ends of the solenoid?