JEE Physics Moving Charges and Magnetism MCQs Set 10

Answer: (a) 8 F1

Explanation:
1. Force between parallel wires: \( F=\frac{\mu_0 I_1 I_2 l}{2\pi d} \).
2. Each current is doubled, so the product I1 I2 becomes 4 times; the distance is halved, so the force is doubled again.
3. \( F_2=4\times 2\,F_1=8F_1 \)

Teacher's Note:
1. Force per unit length between parallel wires is proportional to the product of the currents and inversely proportional to the separation.
2. Treat each change as a separate multiplying factor.

Answer: (c) \( 2\sqrt2:1 \)

Explanation:
1. At the centre: \( B_c=\frac{\mu_0 I}{2r} \).
2. On the axis at distance x: \( B=\frac{\mu_0 I r^2}{2(r^2+x^2)^{3/2}} \). For x = r this gives \( \frac{\mu_0 I r^2}{2(2r^2)^{3/2}}=\frac{\mu_0 I}{4\sqrt2\,r} \).
3. \( \frac{B_c}{B_{axis}}=\frac{1/2}{1/(4\sqrt2)}=2\sqrt2:1 \)

Teacher's Note:
1. The axial field formula reduces to the centre value when x = 0.
2. Always simplify (2r²)^(3/2) = 2√2 r³ before taking ratios.

Answer: (b) \( 176\times10^{-4}\ \text{T} \)

Explanation:
1. Turns per metre: n = 70 x 100 = 7000 per m.
2. \( B=\mu_0 nI=4\pi\times10^{-7}\times7000\times2=1.76\times10^{-2}\ \text{T} \) = 176 x 10⁻⁴ T.

Teacher's Note:
1. Convert turns per cm to turns per metre before using B = mu0 n I.
2. The field inside a long solenoid is uniform and does not depend on its radius.

Answer: (a) \( \frac{9}{130}\ \text{N} \)

Explanation:
1. The field is along the 13 cm side. The angle between the 5 cm side and the 13 cm side satisfies \( \sin\theta=\frac{12}{13} \) (the side opposite to it is 12 cm).
2. \( F=ILB\sin\theta=2\times0.05\times0.75\times\frac{12}{13}=\frac{9}{130}\ \text{N} \)

Teacher's Note:
1. Force on a straight wire in a field is I L B sin(theta), where theta is the angle between the wire and the field.
2. In a 5-12-13 triangle the angle between the 5 side and the hypotenuse has sine 12/13.

Answer: (b) 1.0

Explanation:
1. At equilibrium the magnetic torque balances the restoring torque: \( NIAB=k\phi \).
2. \( A=\frac{k\phi}{NIB}=\frac{4\times10^{-5}\times0.05}{200\times10^{-2}\times0.01}=10^{-4}\ \text{m}^2=1\ \text{cm}^2 \)

Teacher's Note:
1. In a moving coil galvanometer the deflection is proportional to current because N I A B = k phi.
2. Convert mA to A and m² to cm² carefully (1 m² = 10⁴ cm²).

Answer: (b) \( 1.2\times10^{3}\ \text{A m}^{-1} \)

Explanation:
1. Turns per unit length: n = 1200/2 = 600 per m.
2. \( H=nI=600\times2=1.2\times10^{3}\ \text{A m}^{-1} \)

Teacher's Note:
1. Magnetic intensity H = nI is measured in A/m and, unlike B, does not contain mu0.
2. The diameter is not needed because the solenoid is long compared with its width.

Answer: (a) \( 68\times10^{-6}\ \text{T} \)

Explanation:
1. P, with the two wires, forms a right isosceles triangle whose hypotenuse is 7 cm, so the distance from each wire is \( r=\frac{7}{\sqrt2}\approx\frac{7}{1.4}=5\ \text{cm} \).
2. \( B_1=\frac{\mu_0\,8}{2\pi r}=3.2\times10^{-5}\ \text{T},\quad B_2=\frac{\mu_0\,15}{2\pi r}=6\times10^{-5}\ \text{T} \) and the two fields are perpendicular.
3. \( B=\sqrt{B_1^2+B_2^2}=2\times10^{-7}\times\frac{17}{0.05}=68\times10^{-6}\ \text{T} \)

Teacher's Note:
1. The field of each wire at P is perpendicular to the line joining P to that wire, so the two fields are at right angles.
2. 8, 15, 17 is a Pythagorean triple, which makes the resultant easy.

Answer: (b) \( 1.0\times10^{-3}\ \text{J} \)

Explanation:
1. Induced emf: \( \varepsilon=\frac{BA}{t}=\frac{40\times25\times10^{-4}}{1}=0.1\ \text{V} \).
2. Power dissipated: \( P=\frac{\varepsilon^2}{R}=\frac{0.01}{10}=10^{-3}\ \text{W} \).
3. Work done = P t = 1.0 x 10⁻³ J.

Teacher's Note:
1. Pulling the loop out slowly means the work done equals the heat produced in the loop.
2. The flux changes from BA to zero, so the average emf is BA divided by the time.

Answer: (c) \( N_AI_A=4N_BI_B \)

Explanation:
1. Magnetic moment of a coil: \( m=NI\pi r^2 \).
2. Equal moments: \( N_AI_A r_A^2=N_BI_B r_B^2\Rightarrow N_AI_A=\frac{r_B^2}{r_A^2}N_BI_B=4N_BI_B \)

Teacher's Note:
1. The moment of a coil depends on turns, current and the area of one turn.
2. The coil with the smaller radius needs a larger product N I.

Answer: (d) 10 V

Explanation:
1. Field: \( B=10^3\ \text{G}=0.1\ \text{T} \). Only the top side (length 0.1 m) lies in the field.
2. Balance: \( IlB=mg\Rightarrow I=\frac{10^{-3}\times10}{0.1\times0.1}=1\ \text{A} \).
3. \( V=IR=1\times10=10\ \text{V} \)

Teacher's Note:
1. Only the side lying in the field feels a force; the forces on the vertical sides cancel each other.
2. 1 gauss = 10⁻⁴ tesla.

Answer: (d) \( 3\sqrt3\times10^{-5}\ \text{T} \)

Explanation:
1. Distance of the centroid from each side: \( r=\frac{a}{2\sqrt3}=\frac{4\sqrt3}{2\sqrt3}=2\ \text{cm} \).
2. Field of one side: \( B_1=\frac{\mu_0 I}{4\pi r}(\sin60^\circ+\sin60^\circ)=\frac{10^{-7}\times2}{0.02}\times\sqrt3=\sqrt3\times10^{-5}\ \text{T} \).
3. The three sides give fields in the same direction, so \( B=3B_1=3\sqrt3\times10^{-5}\ \text{T} \).

Teacher's Note:
1. For a finite straight wire the field at a point at perpendicular distance r is mu0 I/(4 pi r) times (sin a + sin b).
2. In an equilateral triangle the centroid is equally placed from the three sides and all contributions add.

Answer: (b) 1 : 4

Explanation:
1. The two halves of the loop (through P and Q, and through R and S) are identical, so the current divides equally: each of PQ and RS carries I/2 in the same direction.
2. Force per unit length on PQ due to RS: \( f=\frac{\mu_0 (I/2)(I/2)}{2\pi d}\propto I^2 \).
3. Doubling the reading multiplies f by 4, so the ratio is 1 : 4.

Teacher's Note:
1. Both the current in PQ and the current in RS are proportional to the ammeter reading.
2. So the force between them depends on the square of the reading.

Answer: (a) 4 J

Explanation:
1. Potential energy: \( U=-\vec m\cdot\vec B \). Parallel: U = -mB; antiparallel: U = +mB.
2. \( W=\Delta U=2mB=2\times5.0\times0.4=4\ \text{J} \)

Teacher's Note:
1. Work done by the external agent equals the change in potential energy of the dipole.
2. Turning through 180° always costs 2mB.

Answer: (c) 125

Explanation:
1. Turns per metre: n = 400/0.4 = 1000.
2. Flux: \( \phi=\mu_0\mu_r nIA\Rightarrow \mu_r=\frac{\phi}{\mu_0 nIA} \).
3. \( \mu_r=\frac{4\pi\times10^{-6}}{4\pi\times10^{-7}\times1000\times0.4\times2\times10^{-4}}=125 \)

Teacher's Note:
1. Inside a core of relative permeability mu_r, B = mu_r mu0 n I.
2. Convert the area to m² (2 cm² = 2 x 10⁻⁴ m²).

Answer: (c) \( N^2:n^2 \)

Explanation:
1. Let the wire length be L. For N turns, \( 2\pi rN=L\Rightarrow r=\frac{L}{2\pi N} \).
2. Field at the centre: \( B=\frac{\mu_0NI}{2r}=\frac{\mu_0 N^2\pi I}{L} \), so B is proportional to N².
3. \( \frac{B_1}{B_2}=\frac{N^2}{n^2} \)

Teacher's Note:
1. More turns give both a larger N in the formula and a smaller radius, so B grows as N².
2. The wire length is fixed, which links the radius to the number of turns.

Answer: (d) 0%

Explanation:
1. Current sensitivity: \( \frac{\phi}{I}=\frac{NAB}{k} \), so it is proportional to N. A 50% rise means N becomes 1.5 times.
2. Voltage sensitivity: \( \frac{\phi}{V}=\frac{NAB}{kR} \). The resistance R of the coil is proportional to the length of wire, hence to N.
3. So N/R stays constant and the voltage sensitivity does not change: 0%.

Teacher's Note:
1. Adding turns also adds wire, so the coil resistance increases in the same proportion.
2. This is why simply increasing N does not improve voltage sensitivity.

Answer: (c) \( \frac{\mu_0 i}{2r}\left(\frac12+\frac{1}{2\pi}\right) \)

Explanation:
1. Semicircle: \( B_1=\frac{\mu_0 i}{4r} \), directed out of the page.
2. Upper wire: P lies at perpendicular distance r from the end of a semi-infinite wire, so \( B_2=\frac{\mu_0 i}{4\pi r} \), also out of the page.
3. Lower wire: P lies on the line of the wire, so its contribution is zero.
4. \( B=\frac{\mu_0 i}{4r}+\frac{\mu_0 i}{4\pi r}=\frac{\mu_0 i}{2r}\left(\frac12+\frac{1}{2\pi}\right) \)

Teacher's Note:
1. A semi-infinite wire gives half the field of an infinite wire at a point opposite its end.
2. A point on the line of a straight wire receives no field from that wire.

Answer: (a) \( 144\times10^{-18}\ \text{kg} \)

Explanation:
1. Speed gained: \( qV=\tfrac12mv^2\Rightarrow v=\sqrt{\frac{2qV}{m}} \).
2. Radius in the field: \( r=\frac{mv}{qB}\Rightarrow r^2=\frac{2mV}{qB^2} \).
3. \( m=\frac{qB^2r^2}{2V}=\frac{2\times10^{-6}\times(4\times10^{-3})^2\times(3\times10^{-2})^2}{200}=1.44\times10^{-16}\ \text{kg} \) = 144 x 10⁻¹⁸ kg.

Teacher's Note:
1. Eliminate the speed between the energy equation and the radius equation.
2. Keep every quantity in SI units: microcoulomb, millitesla and centimetre must be converted.

Answer: (d) 3 μT

Explanation:
1. The straight parts lie along a line through O, so they produce no field at O.
2. Semicircle: \( B=\frac{\mu_0 I}{4r}=\frac{4\pi\times10^{-7}\times3}{4\times\pi/10}=3\times10^{-6}\ \text{T} \) = 3 μT.

Teacher's Note:
1. The field at the centre of a semicircular arc is half that of a full circular loop.
2. Wires whose line passes through the point do not contribute.