2. The magnetic susceptibility of silicon is -0.4 × 10^-5. Calculate the flux density and magnetic moment per unit volume when magnetic field of intensity 5 × 10^5 A/m is applied. 3. The magnetic field strength in silicon is 1000 A/m. If the magnetic susceptibility is -0.25 × 10^-5, calculate the magnetization and flux density in silicon. 4. A circular loop of copper having a diameter of 10 cm carries a current of 500 mA. Calculate the magnetic moment associated with the loop. 5. An electron in a hydrogen atom circulates with a radius 0.052 nm. Calculate the change in its magnetic moment if a magnetic induction (B) = 3 Wb/m2 acts at right angles to the plane of orbit.
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4 \times 10^{-5}\) - Magnetic field intensity, \(H = 5 \times 10^5 \, \text{A/m}\) #### Find: - Flux density, \(B\) - Magnetic moment per unit volume, \(M\) #### Show more…
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Timothy J.
'A circular wire loop of radius 0.10 m and resistance 50.09 is suspended horizontally in a magnetic field of magnitude B directed at an angle of 60" with the vertical through the loop; as shown in the figure on the provided sheet: The magnitude of the field, in teslas, is given by the function B(t) = 4(1 0.2 t) where t represents time in seconds. Note that the magnetic field is uniform spatially over the area of the loop (A) Determine an expression for the magnetic flux $B through the loop as function of time. (B) Graph the magnetic flux as a function of time on the axes provided. Determine the magnitude of the induced emf in the loop. Indicate the direction of the induced current on the diagram .'
The quarter-circle loop shown in has an area of $15 \mathrm{~cm}^{2}$. A constant magnetic field, $\mathrm{B}=0.16 \mathrm{~T}$, pointing in the $+\chi$ -direction, fills the space independent of the loop. Find the flux through the loop in each orientation shown. The magnetic flux is determined by the amount of $\overrightarrow{\mathbf{B}}$ -field passing perpendicularly through the particular area, times that area. That is, $\Phi_{M}=B_{\perp} \mathrm{A} .$ (a) $\Phi_{M}=B_{\perp} A=B A=(0.16 \mathrm{~T})\left(15 \times 10^{-4} \mathrm{~m}^{2}\right)=2.4 \times 10^{-4} \mathrm{~Wb}$ (b) $\Phi_{M}=\left(B \cos 20^{\circ}\right) A=\left(2.4 \times 10^{-4} \mathrm{~Wb}\right)\left(\cos 20^{\circ}\right)=2.3 \times 10^{-4} \mathrm{~Wb}$ (c) $\Phi_{M}=\left(B \sin 20^{\circ}\right) A=\left(2.4 \times 10^{-4} \mathrm{~Wb}\right)\left(\sin 20^{\circ}\right)=8.2 \times 10^{-5} \mathrm{~Wb}$
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