Question

The component of magnetic field along the line OP at the point P due to the bar magnet of moment $1.732A{m}^2$ is:

• A. $\sqrt{3}\times {10}^{-7}T$
• B. ${10}^{-7}T$
• C. $3\times {10}^{-7}T$
• D. $23\times {10}^{-7}T$

Answer 1

The correct option is C $3\times {10}^{-7}T$

Given that,

$M=1.732A{m}^2$ Magnetic moment of bar magnet,

Now, the radial component of magnetic field due to bar magnet of magnetic moment is

$B_r=\frac{{\mu }_{0}2M\cos \theta }{4\pi r^3}$

$r$ = magnitude of vector $OP$

${\mu }_0=4\pi {10}^{-7}$

Now, put the value of all elements

$B_r=\frac{4\pi \times {10}^{-7}\times 2M\cos {30}^0}{4\pi \times {\left(1\right)}^3}$

$B_r={10}^{-7}\times 2\times 1.732\times \frac{\sqrt{3}}{2}$

$B_r=2.9\times {10}^{-7}T$

$B_r=3\times {10}^{-7}T$

Hence, the component of magnetic field is $3\times {10}^{-7}T$