JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Magnetic Effects of Current

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A tightly wound long solenoid carries a current of 1.5 A. An electron is executing uniform circular motion inside the solenoid with a time period of 75 ns . The number of turns per metre in the solenoid is ______.
[Take mass of electron

m_{e} = 9 \times 1 0^{- 31} kg

, charge of electron

∣ q_{e} ∣ = 1.6 \times 1 0^{- 19} C

μ_{0} = 4 π \times 1 0^{- 7} \frac{N}{A ^{2}}, 1 ns = 1 0^{- 9} s

]

Answer: 250

Solution

The time period of a charged particle moving in a uniform magnetic field is

T = \frac{2 π m}{q B} .

Inside a long solenoid with

n

turns per metre carrying current

I

, the field is

B = μ_{0} n I .

Combining,

T = \frac{2 π m}{q μ _{0} n I} ⟹ n = \frac{2 π m}{q T μ _{0} I} .

Substituting

m = 9 \times 1 0^{- 31} kg

q = 1.6 \times 1 0^{- 19} C

T = 75 \times 1 0^{- 9} s

μ_{0} = 4 π \times 1 0^{- 7} N/A^{2}

I = 1.5 A

n = \frac{2 π \times 9 \times 1 0 ^{- 31}}{( 1.6 \times 1 0 ^{- 19} ) ( 75 \times 1 0 ^{- 9} ) ( 4 π \times 1 0 ^{- 7} ) ( 1.5 )} .

The

π

cancels between numerator and denominator:

n = \frac{2 \times 9 \times 1 0 ^{- 31}}{1.6 \times 75 \times 4 \times 1.5 \times 1 0 ^{- 35}} = \frac{18 \times 1 0 ^{- 31}}{720 \times 1 0 ^{- 35}} = \frac{18}{720} \times 1 0^{4} = 0.025 \times 1 0^{4} = 250.

The number of turns per metre is

250

JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Integer answer

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JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Magnetic Effects of Current

Solution

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