A solenoid of 1200 turns is wound uniformly in a single layer on a glass tube 2 m long and 0.2 m in diameter. The magnetic intensity at the center of the solenoid when a current of 2 A flows through it is [2023]

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Q.
A solenoid of 1200 turns is wound uniformly in a single layer on a glass tube 2 m long and 0.2 m in diameter. The magnetic intensity at the center of the solenoid when a current of 2 A flows through it is [2023]

1 $2.4 \times 10^{3} {Am}^{- 1}$

2 $1.2 \times 10^{3} {Am}^{- 1}$

3 $1 {Am}^{- 1}$

4 $2.4 \times 10^{- 3} {Am}^{- 1}$

Ans.
(2)

$B = μ_{0} n i = μ_{0} \frac{N}{l} i$

$= 4 π \times 10^{- 7} \times \frac{1200}{2} \times 2 = 48 π \times 10^{- 5} T$

$H = \frac{B}{μ_{0}} = \frac{48 π \times 10^{- 5}}{4 π \times 10^{- 7}} = 12 \times 10^{2} = 1.2 \times 10^{3} A/m$

Solution

Q.
A solenoid of 1200 turns is wound uniformly in a single layer on a glass tube 2 m long and 0.2 m in diameter. The magnetic intensity at the center of the solenoid when a current of 2 A flows through it is [2023]

1 $2.4 \times 10^{3} {Am}^{- 1}$

2 $1.2 \times 10^{3} {Am}^{- 1}$

3 $1 {Am}^{- 1}$

4 $2.4 \times 10^{- 3} {Am}^{- 1}$

Ans.
(2)

$B = μ_{0} n i = μ_{0} \frac{N}{l} i$

$= 4 π \times 10^{- 7} \times \frac{1200}{2} \times 2 = 48 π \times 10^{- 5} T$

$H = \frac{B}{μ_{0}} = \frac{48 π \times 10^{- 5}}{4 π \times 10^{- 7}} = 12 \times 10^{2} = 1.2 \times 10^{3} A/m$

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Solution

Q. A solenoid of 1200 turns is wound uniformly in a single layer on a glass tube 2 m long and 0.2 m in diameter. The magnetic intensity at the center of the solenoid when a current of 2 A flows through it is [2023]

1 2.4×103 Am-1

2 1.2×103 Am-1

3 1 Am-1

4 2.4×10-3 Am-1

Ans. (2) B=μ0ni=μ0Nli =4π×10-7×12002×2=48π×10-5 T H=Bμ0=48π×10-54π×10-7=12×102=1.2×103 A/m

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Q.
A solenoid of 1200 turns is wound uniformly in a single layer on a glass tube 2 m long and 0.2 m in diameter. The magnetic intensity at the center of the solenoid when a current of 2 A flows through it is [2023]

1 $2.4 \times 10^{3} {Am}^{- 1}$

2 $1.2 \times 10^{3} {Am}^{- 1}$

3 $1 {Am}^{- 1}$

4 $2.4 \times 10^{- 3} {Am}^{- 1}$

Ans.
(2)

$B = μ_{0} n i = μ_{0} \frac{N}{l} i$

$= 4 π \times 10^{- 7} \times \frac{1200}{2} \times 2 = 48 π \times 10^{- 5} T$

$H = \frac{B}{μ_{0}} = \frac{48 π \times 10^{- 5}}{4 π \times 10^{- 7}} = 12 \times 10^{2} = 1.2 \times 10^{3} A/m$