Question

The magnetic flux through a coreless solenoid
carrying current I is 5 x 10-6 Wb. The length of
the solenoid is 25 cm, its magnetic moment is
equal to :-
GET PHYENGWODULE OS10T-MAGNE​

Answer 1

Answer:

hear the answer is

Explanation:

Flux = B.dA

where dA is the area associated with the flux.

Therefore,

5*10^6 = 0.25*B

B = 2 x 10^7 T would be the magnetic moment.

Answer 2

The magnetic moment is 0.994 Am².

Explanation:

Given that,

Current $I=5\times10^{-6}\ Wb$

Solenoid = 25 cm

We need to calculate the magnetic moment

Using formula of magnetic flux

$\phi=BA$

$A=\dfrac{\phi}{B}$...(I)

$B = \dfrac{\mu_{0}NI}{L}$

Put the value of B in equation (I)

$A=\dfrac{\phi\times L}{\mu_{0}NI}$

$NIA=\dfrac{\phi\times l}{\mu_{0}}$

Put the value into the formula

$NIA=\dfrac{5\times10^{-6}\times25\times10^{-2}}{4\pi\times10^{-7}}$

$M=NIA=0.994\ Am^2$

Hence, The magnetic moment is 0.994 Am²

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Topic : magnetic moment

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