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  • Verify the Ampere s law for magnetic field of a point dipole of dipole moment

Verify the Ampere’s law for magnetic field of a point dipole of dipole moment m = \widehat{m}k. Take C as the closed curve running clockwise along (i) the z-axis from z = a > 0 to z = R; (ii) along the quarter circle of radius R and center at the origin, in the first quadrant of x-z plane; (iii) along the x-axis from x = R to x = a, and (iv) along the quarter circle of radius an and center at the origin in the first quadrant of x-z plane.


 

Answers (1)

Assume the x-z plane (shown below). All points from P to Q lie on the axial line NS placed at the origin.


The magnetic field at a distance r is 

 B=\frac{\mu _02\left |M \right |}{4\pi r^3}=\frac{\mu_0M}{2\pi r^3 }

Along z-axis from P to Q  

\int_{P}^{Q}B.dl=\int_{P}^{Q} B.dl\cos 0^{\circ}=\int_{a}^{R}Bdz=\int_{a}^{R}\frac{\mu_{0}M}{2\pi r^3}dz=\frac{\mu _0M}{2\pi} \left (-\frac{1}{}2 \right )\left (\frac{1}{R^2}-\frac{1}{a^2} \right )=\frac{\mu _0M}{4\pi} \left (\frac{1}{a^2}-\frac{1}{R^2} \right )

ii Along the quarter circle QS (radius R)


 

Consider point A to lie on the equatorial line of magnetic dipole of moment M sinθ.

Magnetic field at A is 

B_t=\frac{\frac{\mu _0}{4\pi} M\sin\theta}{R^{3}} ;dl=Rd\theta

B_r=\frac{\frac{\mu _0}{2\pi} m\cos \theta}{R^{3}}

   \int_{0}^{\frac{\pi}{2}}B.dl=\int_{0}^{\frac{\pi}{2}} B_tdl\cos 0^{\circ}+ \int_{0}^{\frac{\pi}{2}} B_ r dl\cos 90^{\circ} =\int_{0}^{\frac{\pi}{2}}\frac{\mu _0m}{4\pi R^3}\sin \theta (Rd\theta )=\frac{\mu _{0}m}{4 \pi R^2} \int_{0}^{\frac{\pi}{2}}\sin\theta d\theta =\frac{\mu_{0}m}{4\pi R^2 }

 iii Along x-axis over the path ST, consider the figure given below

From the figure, every point lies on the equatorial line of the magnetic dipole.

 Magnetic field induction at a point distance x from the dipole is

 B=\frac{\frac{\mu _0}{4\pi} M}{x^3 }

\int_{S}^{T}B.dl=\int_{R}^{a}-\frac{\mu _0M}{4\pi x^3} =0 angle between-M and dl is 90^{\circ}

 iv  Along the quarter circle TP of radius a. 

 Let's consider the figure given below 


 

From case ii  we get line integral of B along the quarter circle TP of radius a

 is circular arc TP. 

\int B.dl=\int_{\frac{\pi}{2}}^{0} \frac{\mu _{0}}{4\pi} M\frac{\sin \theta }{a^3}ad\theta =\frac{\mu _0M}{4\pi a^2} \int_{\frac{\pi}{2}}^{0} \sin\theta d\theta =\frac{\frac{\mu _{0}}{4\pi} M}{a^2} \int_{\frac{\pi}{2}}^{0}[-\cos\theta ]

=-\frac{\frac{\mu _0}{4\pi} M}{a^2 }

\int B.dl= \int_{P}^{Q}B.dl+\int_{Q}^{S}B.dl+\int_{S}^{T}B.dl+\int_{T}^{P}B.dl

=\frac{\mu _0M}{4}\left [\frac{1}{a^2}-\frac{1}{R^2} \right ]+\frac{\mu _0M}{4\pi R^2}+0+\left (-\frac{\mu _0M}{4\pi R^2} \right )=0
 

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infoexpert24

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