A magnetizing field of 1600Am produces and magnetic flux of 2.4×10−5weber in a bar of iron of cross-section 0.2×10−4m2 . The peremeability and susceptibililty of the bar is.

μr=μμ0=34×10−34π×10−7=600 μ=ϕAH=2.4×10−50.2×10−4×1600 =7.5×10−4T.A−1m=μr=1+χχ=600−1=599≃596.1

A magnetizing field of 1600Am produces and magnetic flux of 2.4×10−5weber in a bar of iron of cross-section 0.2×10−4m2 . The peremeability and susceptibililty of the bar is.

μr=μμ0=34×10−34π×10−7=600 μ=ϕAH=2.4×10−50.2×10−4×1600 =7.5×10−4T.A−1m=μr=1+χχ=600−1=599≃596.1

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

A magnetizing field of 1600 $\frac{A}{m}$ produces and magnetic flux of $2.4 \times 10^{- 5}$ weber in a bar of iron of cross-section $0.2 \times 10^{- 4} m^{2}$ . The peremeability and susceptibililty of the bar is.

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$7.5 \times 10^{- 5} T A^{- 1} m, 496.1$

$7.5 \times 10^{- 5} T A^{- 1} m, 596.1$

$1.5 \times 10^{- 4} T A^{- 1} m, 596.1$

$1.5 \times 10^{- 4} T A^{- 1} m, 396.1$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{aligned} μ_{r} = \frac{μ}{μ_{0}} = \frac{\frac{3}{4} \times 10^{- 3}}{4 π \times 10^{- 7}} = 600 \\ μ = \frac{ϕ}{A H} = \frac{2.4 \times 10^{- 5}}{0.2 \times 10^{- 4} \times 1600} \\ = 7.5 \times 10^{- 4} T . A^{- 1} m \end{aligned}$
$\begin{array}{ll} = μ_{r} = 1 + χ & χ = 600 - 1 \\ = 599 ≃ 596.1 \end{array}$