(i) Use Biot-Savart law to obtain a formula for magnetic field at the centre of a circular loop of radius R carrying a steady current I. Indicate the direction of the magnetic field. |
(ii) A solenoid of length 5m has a radius of 1 cm and is made of 500 turns it carries a current of 5 A. |
What is the magnitude of the magnetic field inside the solenoid. |
Or |
(i) A short bar magnet of magnetic moment \[=\text{ }32\text{ }A{{m}^{2}}\] is placed in a uniform magnetic field of 15 T. If the bar is free to rotate in the plane of field, which orientations would correspond to its |
(a) stable and |
(b) unstable equilibrium? What is the potential energy of the magnet in each case? |
(ii) State Gauss' law in magnetism. |
(iii) Define magnetic susceptibility of a material. Name two elements, one having positive susceptibility and other having negative susceptibility. |
Answer:
(i) Consider a circular loop of radius R carrying current I. Magnetic field at its centre is given by According to Biot-Savart?s law, the magnetic field at the centre of the loop due to this current element, \[dB=\frac{{{\mu }_{0}}}{4\pi }i\frac{dI\times r}{{{r}^{3}}}\] Since, the direction of \[dI\] is along the tangent, so\[d\,\,I\bot r.\] Thus, \[|d\,\,B|=dB=\frac{{{\mu }_{0}}}{4\pi }\frac{idIr}{{{r}^{3}}}=\frac{{{\mu }_{0}}}{4\pi }\cdot \frac{idI}{{{r}^{2}}}\] \[\therefore \] Magnetic field at pointO due to complete loop, \[\int{dB=B=\int_{0}^{2\pi r}{\frac{{{\mu }_{0}}}{4\pi }\frac{idI}{{{r}^{2}}}}}\] \[=\frac{{{\mu }_{0}}}{4\pi }\cdot \frac{i2\pi r}{{{r}^{2}}}=\frac{{{\mu }_{0}}}{2}\frac{i}{r}\] \[B=\frac{{{\mu }_{0}}}{2}\cdot \frac{i}{r}\] The direction of magnetic field at the centre of circular loop is given by right hand rule. As current carrying loop has the magnetic field lines around it which exerts a force on a moving charge. Thus, it behaves as a magnet with two mutually opposite poles. The anti-clockwise flow of current behaves like a North pole, whereas clockwise flow as South pole. Hence, loop behaves as a magnet. (ii) Given,\[l=5\,\,\text{m},\,\,r=1\,\,\text{cm}\] \[N=500,\,\,I=5A\] \[\therefore \]Number of turns per unit length\[(n)=\frac{N}{l}\] \[n=\frac{500}{5}=100\] Magnetic field inside the solenoid \[(B)={{\mu }_{0}}\,\,n\,\,I\] \[B=4\pi \times 10h-7\times 100\times 5=20\pi \times {{10}^{-5}}\,\,\text{T}\] Or (i) (a) For stable equilibrium Magnetic moment is parallel to B in this condition, the net force and torque on the bar magnet is zero, \[F=0\,\,\text{and}\,\,\tau =0\] and its potential energy U is minimum \[\because \] \[U=-M\cdot B=-MB\cos \theta \] \[\because \] \[\theta =0\] \[\therefore \] \[U=-MB\](minimum)\[=-32\times 15=-480\,\,\text{J}\] (b) For unstable equilibrium Magnetic moment is antiparallel to B, i.e. \[\theta =180{}^\circ \] \[U'=-MB\cos \theta =-32\times 15(-1)=480\,\,\text{J}\] (ii) Gauss?s law in Magnetism According to this law, the net magnetic flux \[({{\phi }_{B}})\] through any closed surface is always zero. i.e. \[{{\phi }_{B}}=\oint{B\cdot dS=0}\] (iii) Magnetic susceptibility It is equal to the ratio of intensity of magnetisation (I) induced in the material to the magnetising force (H). It is denoted by \[{{\chi }_{m}}\]. i.e. \[{{\chi }_{m}}=\frac{I}{H}\] Positive susceptibility \[\to \] paramagnetic substances, like aluminium, sodium. Negative susceptibility \[\to \] Diamagnetic substances, like copper, gold.
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