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Magnetics question

2022 · 25 Jun · Shift 2 · Q56
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Magnetics question

2022 · 25 Jun · Shift 2 · Q56

JEE MainPhysicsMagneticsMCQ+4 / −1
A long solenoid carrying a current produces a magnetic field B along its axis. If the current is doubled and the number of turns per cm is halved, the new value of magnetic field will be equal to
  1. A
    B
  2. B
    2B
  3. C
    4B
  4. D
    B2{B \over 2}2B​
View written solutionFree

Correct answer: A

  1. Magnetic field inside a long solenoid

    The magnetic field inside a long solenoid is given by B=μ0nIB = \mu_0 n IB=μ0​nI where:

    • μ0\mu_0μ0​ = permeability of free space
    • nnn = number of turns per unit length
    • III = current

  2. Initial field

    Initially, B=μ0nIB = \mu_0 n IB=μ0​nI

  3. Changed quantities

    According to the question:

    • current is doubled: I′=2II' = 2II′=2I
    • number of turns per cm is halved: n′=n2n' = \frac{n}{2}n′=2n​

  4. New magnetic field

    Using the same formula, B′=μ0n′I′B' = \mu_0 n' I'B′=μ0​n′I′ Substituting the new values: B′=μ0(n2)(2I)B' = \mu_0 \left(\frac{n}{2}\right)(2I)B′=μ0​(2n​)(2I)

    B′=μ0nIB' = \mu_0 n IB′=μ0​nI

    B′=BB' = BB′=B

  5. Option check

    • A: BBB ✅
    • B: 2B2B2B ❌
    • C: 4B4B4B ❌
    • D: B2\dfrac{B}{2}2B​ ❌

Therefore, the new magnetic field remains unchanged.

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