Magnetic Effects of Current and Magnetism
- A 2 amp current is flowing through two different small circular copper coils having radii ratio 1:2. The ratio of their respective magnetic moments will be Answer: 1. 1:4 Key idea: the magnetic moment of a current loop is M=IA=Iπ r² . The current is the same (2 A), so M∝ r² : M₁/M₂=(r₁/r₂)²=(1/2... 1
- An electron (mass 9×10^-31 kg and charge 1.6×10^-19C) moving with speed c/100 (c = speed of light) is injected into a magnetic field B of magnitude 9×10^-4 T perpendicular to its direction of motion. We wish to apply an uniform electric field E toget… Answer: 2. vec E is perpendicular to vec B and its magnitude is 27×10² V m ⁻¹ Key idea: for the electron to go undeflected, the electric for... 1
- An iron bar of length L has magnetic moment M. It is bent at the middle of its length such that the two arms make an angle 60^° with each other. The magnetic moment of this new magnet is: Answer: 3. M/2 Key idea: Magnetic moment = pole strength × distance between the poles. Bending changes this distance, not the pole strength.... 1
- A tightly wound 100 turns coil of radius 10 cm carries a current of 7 A. The magnitude of the magnetic field at the centre of the coil is (Take permeability of free space as 4π×10^-7 SI units): Answer: 4. 4.4 mT Key idea: Field at the centre of a circular coil of N turns: B=μ₀NI/2R . B=frac4π×10⁻⁷×100×72×0.1=frac8.8×10⁻⁴0.2≈4.4×10⁻³... 1
- Match List-I with List-II.List-I (Material): A. Diamagnetic; B. Ferromagnetic; C. Paramagnetic; D. Non-magneticList-II (Susceptibility ()): I. =0; II. 0>≥-1; III. 1; IV. 0<<ε (a small positive number)Choose the correct answer from the options given b… Answer: 2. A-II, B-III, C-IV, D-I Key idea: The sign and size of magnetic susceptibility chi identify the type of material. Diamagnetic: chi... 1
- In a uniform magnetic field of 0.049 T, a magnetic needle performs 20 complete oscillations in 5 seconds as shown. The moment of inertia of the needle is 9.8×10^-6 kg m^2. If the magnitude of magnetic moment of the needle is x×10^-5 Am^2; then the va… Answer: 1. 1280 π² Key idea: A magnetic needle in a field oscillates like a torsional pendulum: T=2π√(I/MB) . Period T=5/20=0.25 s. Squaring... 1
- A very long conducting wire is bent in a semi-circular shape from A to B as shown in figure. The magnetic field at point P for steady current configuration is given by : Answer: 4. μ₀ i/4R[1-2/π] pointed away from page Add the fields of the three parts at P: two semi-infinite straight wires and one semicircle... 1
- A wire carrying a current I along the positive x-axis has length L. It is kept in a magnetic field B=(2i+3j-4k) T. The magnitude of the magnetic force acting on the wire is : Answer: 4. 5 IL Force on a wire: vec F=I vec L×vec B with vec L=Lhat i . hat i×(2hat i+3hat j-4hat k)=3hat k+4hat j (since hat i×hat j=hat k... 1
- The net magnetic flux through any closed surface is : Answer: 2. Zero Magnetic monopoles do not exist, so field lines always form closed loops: every line entering a closed surface also leaves i... 1
- From Ampere's circuital law for a long straight wire of circular cross-section carrying a steady current, the variation of magnetic field in the inside and outside region of the wire is : Answer: 4. a linearly increasing function of distance r upto the boundary of the wire and then decreasing one with 1/r dependence for the ou... 1
- A long solenoid of radius 1 mm has 100 turns per mm. If 1 A current flows in the solenoid, the magnetic field strength at the centre of the solenoid is : Answer: 3. 12.56 × 10⁻² T Field at the centre of a long solenoid is B=μ₀ nI . n=100 turns/mm =10⁵ turns/m. B=4π×10⁻⁷×10⁵×1=0.1256\ T=12.56×1... 1
- Given below are two statements :Statement I : Biot-Savart's law gives us the expression for the magnetic field strength of an infinitesimal current element (Idl) of a current carrying conductor only.Statement II : Biot-Savart's law is analogous to Co… Answer: 4. Statement I is correct and Statement II is incorrect Statement I: Biot-Savart law gives the field due to a small current element... 1
- A thick current carrying cable of radius 'R' carries current 'I' uniformly distributed across its cross-section. The variation of magnetic field B(r) due to the cable with the distance 'r' from the axis of the cable is represented by : Answer: 3. Graph (3): B vs r, linear rise to a peak then decreasing Key idea: use Ampere's law inside and outside the cable. Inside ( r Outs... 1
- A uniform conducting wire of length 12a and resistance 'R' is wound up as a current carrying coil in the shape of,(i) an equilateral triangle of side 'a'.(ii) a square of side 'a'.The magnetic dipole moments of the coil in each case respectively are… Answer: 1. √(3) Ia² and 3 Ia² Key idea: magnetic moment M=NIA , where N is the number of turns the wire of length 12a gives. Triangle (side... 1
- In the productF = q(v×B) = q v×(Bi+Bj+B_0k)For q = 1 and v = 2i+4j+6k and F = 4i-20j+12kWhat will be the complete expression for B ? Answer: 2. -6i-6j-8k Key idea: expand vec v×vec B as a determinant and match components. Here vec v=(2,4,6) , vec B=(B_x,B_y,B_z) , vec F=(4... 1
- An infinitely long straight conductor carries a current of 5 A as shown. An electron is moving with a speed of 10^5 m/s parallel to the conductor. The perpendicular distance between the electron and the conductor is 20 cm at an instant. Calculate the… Answer: 4. 8×10⁻²⁰ N Key idea: the electron moves parallel to the wire, so vec v⊥vec B and F=evB . B=μ₀I/2π r=dfrac2×10⁻⁷×50.2=5×10⁻⁶\ mathr... 1
- A long solenoid of 50 cm length having 100 turns carries a current of 2.5 A. The magnetic field at the centre of the solenoid is :(μ_0 = 4π×10^-7 T m A^-1) Answer: 1. 6.28×10⁻⁴ T Field inside a long solenoid: B=μ₀nI , with n turns per metre. n=100/0.5=200\ m⁻¹ B=4π×10⁻⁷×200×2.5=6.28×10⁻⁴\ T Answ... 1
- An iron rod of susceptibility 599 is subjected to a magnetising field of 1200 A m^-1. The permeability of the material of the rod is :(μ_0 = 4π×10^-7 T m A^-1) Answer: 1. 2.4π×10⁻⁴ T m A ⁻¹ Relative permeability μ_r=1+chi=1+599=600 . μ=μ₀μ_r=4π×10⁻⁷×600=2.4π×10⁻⁴\ T m A⁻¹ The magnetising field 1200... 1
- A cylindrical conductor of radius R is carrying a constant current. The plot of the magnitude of the magnetic field, B with the distance, d, from the centre of the conductor, is correctly represented by the figure : Answer: 3. [graph] B increasing linearly for d < R, peak at R, then decreasing for d > R By Ampere's law for a cylindrical conductor with un... 1
- At a point A on the earth's surface the angle of dip, δ = +25^°. At a point B on the earth's surface the angle of dip, δ = -25^°. We can interpret that : Answer: 3. A is located in the northern hemisphere and B is located in the southern hemisphere. The angle of dip is positive (the north pole... 1
- Ionized hydrogen atoms and α-particles with same momenta enters perpendicular to a constant magnetic field, B. The ratio of their radii of their paths r_H : r_α will be : Answer: 1. 2 : 1 In a magnetic field a charged particle moves in a circle of radius r=p/qB . Momentum p and B are the same, so r∝ 1/q . Ioni... 1
- A metallic rod of mass per unit length 0.5 kg m^-1 is lying horizontally on a smooth inclined plane which makes an angle of 30^° with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of induct… Answer: 4. 11.32 A For the rod to stay at rest on a smooth incline, the horizontal magnetic force must balance the component of weight down... 1
- A thin diamagnetic rod is placed vertically between the poles of an electromagnet. When the current in the electromagnet is switched on, then the diamagnetic rod is pushed up, out of the horizontal magnetic field. Hence the rod gains gravitational po… Answer: 1. the current source A static magnetic field exerts a force perpendicular to the motion of charges, so it does no net work itself.... 1
- Current sensitivity of a moving coil galvanometer is 5 div/mA and its voltage sensitivity (angular deflection per unit voltage applied) is 20 div/V. The resistance of the galvanometer is Answer: 2. 250 Ω Current sensitivity I_s=θ/I and voltage sensitivity V_s=θ/V . Since V=IG , V_s=I_s/G . G=I_s/V_s=5\ div/mA/20\ div/V=frac5\... 1