Question

what is the magnitude of the equatorial and axial field due to a bar magnet of length? 5cm at a distance of 50cm from its mid pt. ? magnetic moment is 0.40Amsquare

Answer 1

Concept :- magnetic field due to bar magnet of length 2l and having magnetic dipole moment M at distance r from its centre ,
At a point on its axial line , B = $\bold{\frac{\mu_0}{4\pi}\frac{2Mr}{(r^2-l^2)^2}}$
at point on its equatorial line , B = $\bold{\frac{\mu_0}{4\pi}\frac{M}{(r^2+l^2)^{3/2}}}$

Now, come to the question ,
M = 0.4 Am²
2l = 5cm = 5 × 10⁻² m
r = 50 cm = 0.5 m

Now, magnetic field on axial line , B = $\bold{\frac{\mu_0}{4\pi}\frac{2\times0.4\times0.5}{(0.5^2-0.05^2)^2}}$
∵ r >> l so, (r² - l²) ≈ r² e.g., (0.5² - 0.005²) ≈ 0.5²
B = $\bold{\frac{\mu_0}{4\pi}\frac{2\times0.4}{0.5^3}}=\bold{\frac{\mu_0}{4\pi}\frac{0.8}{0.125}}$

similarly, magnetic field on equatorial line , B' = $\bold{\frac{\mu_0}{4\pi}\frac{0.4}{(0.5^2+0.05^2)^{3/2}}}$

Answer 2

Answer:

Explanation:magnetic field due to bar magnet of length 2l and having magnetic dipole moment M at distance r from its centre ,

At a point on its axial line , B =

at point on its equatorial line , B =

Now, come to the question ,

M = 0.4 Am²

2l = 5cm = 5 × 10⁻² m

r = 50 cm = 0.5 m

Now, magnetic field on axial line , B =

∵ r >> l so, (r² - l²) ≈ r² e.g., (0.5² - 0.005²) ≈ 0.5²

B =

similarly, magnetic field on equatorial line