JEE Main · 2021 · 17 Mar 2021 · Shift 1 · Magnetic Effects of Current

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A solenoid of 1000 turns per meter has a core with relative permeability 500 . Insulated windings of the solenoid carry an electric current of 5 A . The magnetic flux density produced by the solenoid is: (permeability of free space

= 4 π \times 1 0^{- 7} H / m

)

(A) $2 \times 1 0^{- 3} π T$
(B) $π T$ ✓ Correct
(C) $1 0^{- 4} π T$
(D) $\frac{π}{5} T$

Answer: (B)

Solution

Inside a long solenoid wound on a core of relative permeability

μ_{r}

B = μ_{0} μ_{r} n I

Substituting

n = 1000

turns per metre,

μ_{r} = 500

and

I = 5

B = (4 π \times 1 0^{- 7}) (500) (1000) (5) = 4 π \times 1 0^{- 7} \times 2.5 \times 1 0^{6} = π T

The magnetic flux density produced by the solenoid is

π

JEE Main · 2021 · 17 Mar 2021 · Shift 1 · Single correct option

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