The magnetic flux near the axis and inside the air core solenoid of length 60 cm carrying current ‘I’ is 1.57 × 10-6 Wb. Its magnetic moment will be (cross-sectional area of a solenoid is very small as compared to its length, μ0 = 4 π × 10- 7 SI unit)

Question

The magnetic flux near the axis and inside the air core solenoid of length 60 cm carrying current ‘I’ is 1.57 × 10-6 Wb. Its magnetic moment will be (cross-sectional area of a solenoid is very small as compared to its length, μ0 = 4 π × 10- 7 SI unit) (A)  0.25 A (B)  0.50 A (C)  0.75 A (D)  1 A

Tardigrade Admin · Accepted Answer

Magnetic induction inside the solenoid B =(μ0 NI / L ) Magnetic Flux,φ= BA =(μ0 NIA / L ) Magnetic moment = NIA =(φ L /μ0) =(1.57 × 10-6 × 0.6/4 × 3.141 × 10-7) =0.75 Am 2

Q. The magnetic flux near the axis and inside the air core solenoid of length $60 c m$ carrying current ‘I’ is $1.57 \times 1 0^{- 6} Wb$ . Its magnetic moment will be (cross-sectional area of a solenoid is very small as compared to its length, $μ_{0} = 4 π \times 1 0^{- 7}$ SI unit)

$0.25 A$

$0.50 A$

$0.75 A$

$1 A$

Magnetic induction inside the solenoid
$B = \frac{μ _{0} N I}{L}$
Magnetic Flux, $ϕ = B A = \frac{μ _{0} N I A}{L}$
Magnetic moment $= N I A = \frac{ϕ L}{μ _{0}}$
$= \frac{1.57 \times 1 0 ^{- 6} \times 0.6}{4 \times 3.141 \times 1 0 ^{- 7}}$
$= 0.75 A m^{2}$

Q. The magnetic flux near the axis and inside the air core solenoid of length 60cm carrying current ‘I’ is 1.57×10−6Wb. Its magnetic moment will be (cross-sectional area of a solenoid is very small as compared to its length, μ0​=4π×10−7 SI unit)

0.25A

0.50A

0.75A

1A