A rod of ferromagnetic material with dimensions 10 cm×0.5 cm×0.2 cm is placed in a magnetic field of strength 0.5 cm×104 A m-1 as a result of which a magnetic moment of 5 A m-2 is produced in the rod. The value of magnetic induction will be

Here, V=10×0.5×0.2 cm3=10-6 m3 H=0.5×104 A m-1=5×103Am-1 M=5 A m-2; B=? I=MV=510-6=5×106 Am-1 B=μ0(H+I)=4π×10-7×5×103+5×106 =4π×10-7×5×106×10-3+1=6.28 T

A rod of ferromagnetic material with dimensions 10 cm×0.5 cm×0.2 cm is placed in a magnetic field of strength 0.5 cm×104 A m-1 as a result of which a magnetic moment of 5 A m-2 is produced in the rod. The value of magnetic induction will be

Here, V=10×0.5×0.2 cm3=10-6 m3 H=0.5×104 A m-1=5×103Am-1 M=5 A m-2; B=? I=MV=510-6=5×106 Am-1 B=μ0(H+I)=4π×10-7×5×103+5×106 =4π×10-7×5×106×10-3+1=6.28 T

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 rod of ferromagnetic material with dimensions $10 cm \times 0.5 cm \times 0.2 cm$ is placed in a magnetic field of strength $0.5 cm \times 10^{4} A m^{- 1}$ as a result of which a magnetic moment of $5 A m^{- 2}$ is produced in the rod. The value of magnetic induction will be

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

$0.36 T$

$6.28 T$

$0.54 T$

$2.52 T$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$Here, V = 10 \times 0.5 \times 0.2 {cm}^{3} = 10^{- 6} m^{3} H = 0.5 \times 10^{4} A m^{- 1} = 5 \times 10^{3} {Am}^{- 1} M = 5 A m^{- 2}; B = ? I = \frac{M}{V} = \frac{5}{10^{- 6}} = 5 \times 10^{6} {Am}^{- 1} B = μ_{0} (H + I) = 4 π \times 10^{- 7} \times (5 \times 10^{3} + 5 \times 10^{6}) = 4 π \times 10^{- 7} \times 5 \times 10^{6} \times (10^{- 3} + 1) = 6.28 T$