Two identical magnetic dipoles of magnetic mo | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(b)With respect to $1^{st} $ magnet, $P$ lies in end side-on position
$	herefore {B_1} = \frac{{{\mu _0}}}{{4\pi }}\left( {\frac{{2M}}{{{d^3}}}} i

Vedclass · Accepted Answer

$\sqrt 5 	imes {10^{ - 7}}\,T$

Vedclass · Answer

(b)With respect to $1^{st} $ magnet, $P$ lies in end side-on position
$	herefore {B_1} = \frac{{{\mu _0}}}{{4\pi }}\left( {\frac{{2M}}{{{d^3}}}} ight)$ $(RHS)$
With respect to $2^{nd} $ magnet. $P$ lies in broad side on position.
$	herefore \;{B_2} = \frac{{{\mu _0}}}{{4\pi }}\left( {\frac{M}{{{d^3}}}} ight)$ (Upward)
${B_1} = {10^{ - 7}} 	imes \frac{{2 	imes 1}}{1} = 2 	imes {10^{ - 7}}T,\;{B_2} = \frac{{{B_1}}}{2} = {10^{ - 7}}\,T$
As $B_1$ and $B_2$ are mutually perpendicular, hence the resultant magnetic field
${B_R} = \sqrt {B_1^2 + B_2^2} = \sqrt {{{(2 	imes {{10}^{ - 7}})}^2} + {{({{10}^{ - 7}})}^2}} $$ = \sqrt 5 	imes {10^{ - 7}}\,T$

Two identical magnetic dipoles of magnetic moments $1.0 \,A-m^2$ each, placed at a separation of $2\,m$ with their axis perpendicular to each other. The resultant magnetic field at a point midway between the dipoles is

Similar Questions

A short bar magnet has a magnetic moment of $0.48\; J \;T ^{-1} .$ Give the direction and magnitude of the magnetic field produced by the magnet at a distance of $10 \,cm$ from the centre of the magnet on

$(a)$ the axis,

$(b)$ the equatorial lines (normal bisector) of the magnet.

The mid points of two small magnetic dipoles of length $d$ in end-on positions, are separated by a distance $x, (x > > d)$. The force between them is proportional to $x^{-n}$ where $n$ is

Magnetic moments of two bar magnets may be compared with the help of

Magnetic lines of force

Of the following Fig., the lines of magnetic induction due to a magnet $SN$, are given by