Question

5. A short bar magnet of magnetic moment
5√3x10^-2 JT- is placed with its axis normal to
the earth's horizontal magnetic field. The distance
from the centre of magnet, at its equatorial position
where the resultant field makes an angle 60° with
earth's horizontal field is (B = 0.4 x 10^-4T)


1)5(3)^1/3 cm

2) 5 cm

3) 5(6)^1/3 cm

4)10 cm
​

Answer 1

Answer- above answer is correct

option-2)

5 cm

Answer 2

Answer:

The distance from the centre of magnet is 5cm i.e.option(2).

Explanation:

From the tangent law we have,

$tan\theta=\frac{B}{B_H}$     (1)

Where,

θ=angle made with the earths horizontl

B=actual magnetic field

$B_H$=magnetic field along horizontal direction

And from the question we have,

$M=5\sqrt{3} \times 10^-^2$

$B_H$=0.4×10⁻⁴

θ=60°

By substituting the required values in equation (1) we get;

$B=B_H\times tan60\textdegree$

$B=0.4\times 10^-^4\times\sqrt{3}$     (2)

And the actual magnetic field in terms of magnetic moment is given as,

$B=\frac{\mu_0\times M}{4\pi \times r^3}$          (3)

By substituting the values and using the equation (2) in equation (3) we get;

$r^3=\frac{4\pi\times 10^-^7\times 5\sqrt{3} \times 10^-^2}{4\pi\times 0.4\times 10^-^4\times\sqrt{3} }$

$r^3=125\times 10^-^6$

$r=5\times 10^-^2 meter$

$r=5cm$             (∵10⁻²m=1cm)

Hence, the correct answer among the given option is 5cm i.e.option(2).