Magnetic field | Electricity and Magnetism | Class 12 NEB Physics | Numerical Problems Discussion
Magnetic Effects of Current Problems
Physics · Electromagnetism, Galvanometers & Fields
Two galvanometers, which are otherwise identical, are fitted with different coils. One has a coil of 50 turns and resistance 10 Ω while the other has 500 turns and a resistance of 600 Ω. What is the ratio of the deflection when each is connected in turns to a cell of emf 25 V and internal resistance 50 Ω?
Coil 2: N2 = 500 turns, R2 = 600 Ω
Cell: E = 25 V, r = 50 Ω
Let \(\alpha_1\) and \(\alpha_2\) be the deflections. At equilibrium, restoring torque equals deflecting torque:
\[k \alpha = BINA \implies I = \frac{k \alpha}{B N A}\]Thus, we can construct expressions for current in both setups:
\[I_1 = \frac{k \alpha_1}{B N_1 A} = \frac{E}{R_1 + r} \quad \text{and} \quad I_2 = \frac{k \alpha_2}{B N_2 A} = \frac{E}{R_2 + r}\]Dividing these conditions yields:
\[\begin{align*} \frac{\alpha_1 N_2}{N_1 \alpha_2} &= \frac{R_2 + r}{R_1 + r}\\ \frac{\alpha_1 \times 500}{50 \times \alpha_2} &= \frac{600 + 50}{10 + 50}\\ 10 \times \frac{\alpha_1}{\alpha_2} &= \frac{650}{60}\\ \therefore \frac{\alpha_1}{\alpha_2} &= \frac{13}{12} \end{align*}\]A silver wire has 1×1030 free electrons per cubic meter, a cross sectional area of 2 mm2 and carries a current of 5 A. Calculate the force acting on each electron if the wire is now placed in a magnetic field of flux density 0.15 T which is perpendicular to the wire.
Current (I) = 5 A | Field (B) = 0.15 T | Angle (\(\theta\)) = 90°
The magnetic force acting on a moving charge carrier is:
\[F_e = B e v_d\]Relating the current density to drift velocity via \(I = neAv_d\):
\[\begin{align*} v_d &= \frac{I}{neA} = \frac{5}{10^{30} \times 1.6 \times 10^{-19} \times 2 \times 10^{-6}} = 1.5625 \times 10^{-5} \text{ m/s}\\ F_e &= 0.15 \times 1.6 \times 10^{-19} \times 1.5625 \times 10^{-5}\\ \therefore F_e &= 3.75 \times 10^{-25} \text{ N} \end{align*}\]A horizontal straight wire 5 cm long weighing 1.2 g/m is placed perpendicular to a uniform horizontal magnetic field of flux density of 0.6 T. If the resistance per unit length of the wire is 3.8 Ω/m, calculate the p.d. that has to be applied between the ends of the wire to make it self-supporting.
Field (B) = 0.6 T | Linear Resistance (\(R/l\)) = 3.8 Ω/m \(\implies R = 0.19\ \Omega\)
For balance, the upward magnetic force must strictly counter gravity (\(F_m = F_g\)):
\[\begin{align*} BIl &= mg \implies I = \left(\frac{m}{l}\right) \frac{g}{B}\\ I &= 1.2 \times 10^{-3} \times \frac{9.8}{0.6} = 0.0196 \text{ A}\\ \end{align*}\]Applying Ohm's law to discover the essential potential difference across the dynamic terminals:
\[V = I \times R = 0.0196 \times 0.19 \approx 3.72 \times 10^{-3} \text{ V}\]Note: If calculated using \(g = 10\text{ m/s}^2\), \(I = 0.02\text{ A}\) and \(V = 3.8 \times 10^{-3}\text{ V}\).
A coil consisting of 100 circular loops with radius 60 cm carries a current of 5 A. Find the magnetic field at a point along the axis of the coil, 80 cm from the centre. (\(\mu_0 = 4\pi\times10^{-7}\) T·m/A)
The axial formulation for a magnetic vector configuration reads:
\[\begin{align*} B &= \frac{\mu_0 n I a^2}{2(a^2 + x^2)^{3/2}}\\ B &= \frac{4\pi \times 10^{-7} \times 100 \times 5 \times (0.60)^2}{2[(0.60)^2 + (0.80)^2]^{3/2}}\\ &= \frac{2.2619 \times 10^{-4}}{2 \times (1.0)^{3/2}} = 1.13 \times 10^{-4} \text{ T} \end{align*}\]A straight conductor of length 5 cm carries current of 1.5 A. The conductor experiences a magnetic force of 4.5×10-3 N when it is placed in a magnetic field of 0.9 T. What angle the conductor makes with magnetic field?
Using the fundamental cross-product mapping property:
\[\begin{align*} F &= BIl\sin\theta\\ 4.5 \times 10^{-3} &= 0.9 \times 1.5 \times 0.05 \times \sin\theta\\ \sin\theta &= \frac{4.5 \times 10^{-3}}{0.0675} = \frac{1}{15}\\ \theta &= \sin^{-1}\left(\frac{1}{15}\right) \approx 3.82^\circ \end{align*}\]A horizontal wire, of length 5 cm and carrying a current of 2 A, is placed in the middle of a long solenoid at right angles to its axis. The solenoid has 1000 turns per meter and carries a steady current I. Calculate I if the force on the wire is equal to 10-4 N. (\(\mu_0 = 4\pi\times10^{-7}\) H/m).
The interior magnetic layout lines of a long solenoid run strictly down its axis. The test wire sits perpendicular to this internal coordinate matrix, creating a perfect orthogonal structure (\(\theta = 90^\circ\)):
\[F = B_{\text{sol}} I_{\text{wire}} l \implies 10^{-4} = B_{\text{sol}} \times 2 \times 0.05 \implies B_{\text{sol}} = 10^{-3} \text{ T}\]Now relate this core parameter back to Ampere's law for solenoids:
\[\begin{align*} B_{\text{sol}} &= \mu_0 n I \implies 10^{-3} = 4\pi \times 10^{-7} \times 1000 \times I\\ I &= \frac{1}{4\pi} \approx 0.796 \text{ A} \end{align*}\]The coil of a moving coil galvanometer has 50 turns and its resistance is 10 Ω. It is replaced by a coil having 100 turns and resistance 50 Ω. Find the factor by which the current and voltage sensitivities change.
The operational definitions for structural measurement sensitivity yield:
\[\text{Current Sensitivity } (I_s) = \frac{NBA}{k} \implies \frac{I_{s2}}{I_{s1}} = \frac{N_2}{N_1} = \frac{100}{50} = 2\] \[\text{Voltage Sensitivity } (V_s) = \frac{NBA}{kR} \implies \frac{V_{s2}}{V_{s1}} = \frac{N_2}{N_1} \times \frac{R_1}{R_2} = \left(\frac{100}{50}\right) \times \left(\frac{10}{50}\right) = \frac{2}{5}\]A 60 cm long wire of mass 10 g is suspended horizontally in a transverse magnetic field of flux density 0.4 T through two springs at its two ends. Calculate the current required to pass through the wire so that there is zero tension in the springs.
To eliminate mechanical load stresses inside the support springs, the total up-directed magnetic lifting thrust force matches the downward gravitational vector:
\[\begin{align*} BIl &= mg\\ 0.4 \times I \times 0.60 &= 0.01 \times 9.8\\ 0.24 \times I &= 0.098\\ \therefore I &\approx 0.408 \text{ A} \end{align*}\]Two long parallel conductors carry respectively currents of 12 A and 8 A in the same direction. If the wires are 10 cm apart, find where a third parallel wire also carrying a current must be placed so that the force experienced by it will be zero.
Let the third wire be positioned between the two original lines at distance \(x\) from the 12 A conductor. Equilibrium requires balancing forces:
\[\begin{align*} F_1 &= F_2 \implies \frac{\mu_0 I_1 I_3}{2\pi x} = \frac{\mu_0 I_2 I_3}{2\pi (0.10 - x)}\\ \frac{12}{x} &= \frac{8}{0.10 - x} \implies 1.2 - 12x = 8x\\ 20x &= 1.2 \implies x = 0.06 \text{ m} = 6 \text{ cm} \end{align*}\]An electron of K.E. 10 eV is moving in a circular orbit of radius 11 cm, in a plane at right angles to a uniform magnetic field. Determine the value of the flux density. (mass of an electron = 9.1×10-31 kg, e = 1.6×10^{-19} C).
Equating the centripetal necessity to magnetic force yields:
\[Bev = \frac{mv^2}{r} \implies B = \frac{mv}{er} = \frac{\sqrt{2m(\text{KE})}}{er}\]Plugging the variables directly into the combined kinetic energy profile:
\[\begin{align*} v &= \sqrt{\frac{2 \times 16 \times 10^{-19}}{9.1 \times 10^{-31}}} \approx 1.876 \times 10^{6} \text{ m/s}\\ B &= \frac{9.1 \times 10^{-31} \times 1.876 \times 10^6}{1.6 \times 10^{-19} \times 0.11} \approx 9.7 \times 10^{-5} \text{ T} \end{align*}\]A copper wire 28 m long is wound into a flat circular coil 8.0 cm in diameter. If the current of 4.50 A flows through the coil, what is the magnetic induction at the centre?
Determine the total number of turns \(n\) from wire constraints:
\[n \times 2\pi r = l \implies n = \frac{28}{2\pi \times 0.04} \approx 111.4 \text{ turns}\]Compute field magnitude at coil center:
\[B = \frac{\mu_0 n I}{2r} = \frac{4\pi \times 10^{-7} \times 111.4 \times 4.50}{2 \times 0.04} \approx 7.87 \times 10^{-3} \text{ T}\]An alpha particle makes a full rotation in a circle of radius 1.0 meter in 2.0 sec. Calculate the value of magnetic field induction at the center of the circle. (\(\mu_0 = 4\pi\times10^{-7}\) H/m)
The convective current created by the orbital charge cycle is:
\[I = \frac{q}{T} = \frac{3.2 \times 10^{-19}}{2.0} = 1.6 \times 10^{-19} \text{ A}\]Calculating field induction at the geometric path center:
\[B = \frac{\mu_0 I}{2r} = \frac{4\pi \times 10^{-7} \times 1.6 \times 10^{-19}}{2 \times 1.0} \approx 1.005 \times 10^{-25} \text{ T}\]A coil consisting of 100 circular loops with radius 0.60 m carries a current of 5 A. At what distance from the center, along the axis, the magnetic field magnitude 1/8 as great as it is at the center?
Write out structural expressions for both spatial configurations:
\[\frac{\mu_0 n I a^2}{2(a^2 + x^2)^{3/2}} = \frac{1}{8} \left( \frac{\mu_0 n I}{2a} \right) \implies \frac{a^2}{(a^2+x^2)^{3/2}} = \frac{1}{8a}\]Rearranging the algebraic equation simplifies evaluation:
\[\begin{align*} 8a^3 &= (a^2+x^2)^{3/2} \implies (2a)^3 = \left(\sqrt{a^2+x^2}\right)^3\\ 2a &= \sqrt{a^2+x^2} \implies 4a^2 = a^2 + x^2\\ x^2 &= 3a^2 \implies x = a\sqrt{3}\\ x &= 0.60 \times 1.732 \approx 1.039 \text{ m} \end{align*}\]
Comments
Post a Comment