JEE Physics Moving Charges and Magnetism MCQs Set 07

Answer: (b) \( \frac{\mu_0 i}{2\pi}\left\{\frac{1}{r-x}+\frac1x\right\} \)

Explanation:
1. For opposite currents the fields of the two wires point the same way in the region between them, so they add.
2. Distances are x and (r - x): \( B=\frac{\mu_0 i}{2\pi}\left\{\frac1x+\frac1{r-x}\right\} \).

Teacher's Note:
1. Between antiparallel currents the fields reinforce.
2. Outside the pair, they would partially cancel.

Answer: (b) zero

Explanation:
1. The coils are identical and the point is midway, so each gives a field of the same magnitude on the axis.
2. The currents are opposite, so the two axial fields are opposite in direction and cancel: B = 0.

Teacher's Note:
1. Opposite currents make the two axial fields opposite in direction.
2. With equal magnitude by symmetry, the resultant is zero.

Answer: (b) a curve (parabola) rising from the origin up to R, then constant

Explanation:
1. For r < R the enclosed current is \( i\frac{r^2}{R^2} \), so \( \oint\vec B\cdot d\vec l=\mu_0 i\frac{r^2}{R^2} \), which grows as \( r^2 \).
2. For r > R the enclosed current is the full i, so the circulation is constant at \( \mu_0 i \).
3. This is the graph (b) of the figure: a parabola up to R and then a horizontal line.

Teacher's Note:
1. The circulation depends on the enclosed current, not on B itself.
2. Enclosed current grows as the enclosed area, i.e. as r squared.

Answer: (b) the magnetic field at any point inside the pipe is zero

Explanation:
1. Take an amperian circle inside the pipe: no current is enclosed, so \( B\cdot2\pi r=0 \).
2. Thus B = 0 at every interior point.

Teacher's Note:
1. A thin-walled pipe carries current only on its surface.
2. Any coaxial amperian loop inside encloses zero current.

Answer: (a) outside the cable

Explanation:
1. Outside the cable the net enclosed current is \( I-I=0 \), so B = 0.
2. Inside, between the conductors, the enclosed current is I, giving a non-zero field.

Teacher's Note:
1. Opposite equal currents cancel in the total enclosed current outside.
2. This is why coaxial cables do not radiate magnetic fields.

Answer: (b) West

Explanation:
1. Take East = +x, North = +y, up = +z. The velocity is \( -\hat k \) (downward) and \( \vec B\propto\hat j \).
2. \( \vec v\times\vec B\propto(-\hat k)\times\hat j=+\hat i \) (East), but the charge is negative, so the force is towards West.

Teacher's Note:
1. For a negative charge reverse the direction obtained from v x B.
2. The force is perpendicular to both the downward velocity and the northward field.

Answer: (d) a helix with increasing pitch

Explanation:
1. The magnetic force makes the particle circle in the plane perpendicular to the fields.
2. The electric force along the common field direction accelerates the particle along the axis, so the distance moved along the axis in one turn increases with time.
3. So the path is a helix of increasing pitch.

Teacher's Note:
1. Pitch = (speed along the field) x (period). The period is constant, but the axial speed keeps increasing.
2. If E were absent the helix would have uniform pitch.

Answer: (b) with larger radii for the faster electrons

Explanation:
1. \( r=\frac{mv}{eB}\propto v \) for fixed B, m and e.
2. So faster electrons move on circles of larger radii.

Teacher's Note:
1. Radius is directly proportional to speed.
2. The period 2 pi m / eB is the same for all speeds.

Answer: (b) negative z-axis

Explanation:
1. Electric force on the electron: \( -e\vec E \) is along +y (E is along -y).
2. The magnetic force must therefore be along -y. For an electron \( \vec F_m=-e\,\vec v\times\vec B \). With \( \vec B=-B\hat k \): \( \vec v\times\vec B\propto\hat i\times(-\hat k)=+\hat j \), and the negative charge reverses it to \( -\hat j \).
3. So B must be along the negative z-axis.

Teacher's Note:
1. For an electron both E-force and B-force have the opposite sign to the positive-charge rule, but only the relative directions matter.
2. Verify with the cross product i x (-k) = +j; times the negative charge gives -j.

Answer: (a) \( y=x \)

Explanation:
1. In the first and third quadrants the two fields are opposite (one into the page, one out of it); in the second and fourth quadrants they are in the same direction.
2. Magnitudes are equal when the distances are equal, i.e. \( |x|=|y| \).
3. Both conditions hold on \( y=x \).

Teacher's Note:
1. Check the direction of each wire's field in each quadrant first.
2. Equal distance from both axes means |x| = |y|.

Answer: (b) 1 J

Explanation:
1. \( W=2MB=2\times2.5\times0.2=1\ \text{J} \).

Teacher's Note:
1. Work to turn a dipole by 180 degrees is 2 M B.
2. This equals the change from U = -MB to U = +MB.

Answer: (b) \( \frac{qvr}{2} \)

Explanation:
1. Current \( I=\frac{q}{T}=\frac{qv}{2\pi r} \).
2. Moment \( =IA=\frac{qv}{2\pi r}\cdot\pi r^2=\frac{qvr}{2} \).

Teacher's Note:
1. A revolving charge gives I = q/T = qv/2 pi r.
2. Multiply by the area pi r^2.

Answer: (b) A and C, and also B and D

Explanation:
1. The field depends only on the perpendicular distance from the x-axis: \( d=\sqrt{y^2+z^2} \).
2. A: 1, B: \( \sqrt2 \), C: 1, D: \( \sqrt2 \).
3. So A and C have equal fields, and B and D have equal fields.

Teacher's Note:
1. The x-coordinate does not matter for an infinite straight wire along x.
2. Compare sqrt(y^2 + z^2) for each point.

Answer: (a) I and IV only

Explanation:
1. Distance of D from the x-axis: \( \sqrt{1^2+1^2}=\sqrt2 \), so \( B=\frac{\mu_0 i}{2\pi\sqrt2} \) (I true, II false).
2. B is perpendicular to the wire, so it is not parallel to x (III false).
3. The field direction is perpendicular to the radial vector \( (0,1,1)/\sqrt2 \) and to the x-axis, i.e. along \( (0,-1,1)/\sqrt2 \), which makes 45° with the xy-plane (IV true).

Teacher's Note:
1. The field of a straight wire circles the wire and is always perpendicular to the plane containing wire and point.
2. Radial direction is at 45 degrees in the yz-plane, and so is the field.

Answer: (b) is constant (zero) inside the tube, is zero at the axis, and is non-zero outside at finite distances

Explanation:
1. By Ampere's law, a circle inside the tube encloses no current, so B = 0 everywhere inside (constant and zero, including the axis).
2. Outside, B = mu0 I/2 pi r, non-zero at finite distances.

Teacher's Note:
1. Inside a uniform thin cylindrical shell of current the field is zero, not linear.
2. The linear rise applies to a solid cylinder.

Answer: (a) abc = 1

Explanation:
1. Time period \( T=\frac{2\pi m}{qB} \) does not depend on the angle: a = 1.
2. Radius \( r\propto v\sin\theta \): \( b=\frac{\sin30^\circ}{\sin60^\circ}=\frac1{\sqrt3} \).
3. Pitch \( \propto v\cos\theta \): \( c=\frac{\cos30^\circ}{\cos60^\circ}=\sqrt3 \).
4. So \( abc=1\times\frac1{\sqrt3}\times\sqrt3=1 \) (also a = bc).

Teacher's Note:
1. Period is the same for both; radius follows sin(theta) and pitch follows cos(theta).
2. The product bc comes out exactly 1.

Answer: (c) III and IV only

Explanation:
1. (I) false: with zero velocity the magnetic force is zero.
2. (II) false: the force does no work, so energy stays constant.
3. (III) true: the force depends on \( B\sin\theta \).
4. (IV) true: \( \vec F\perp\vec v \).

Teacher's Note:
1. A magnetic force needs motion and never changes kinetic energy.
2. The force is v x B, so it is perpendicular to v.

Answer: (b) II and IV only

Explanation:
1. \( r=\frac{mv}{qB}\propto\frac1q \), so \( r_1:r_3=3:1 \) (II true, I false).
2. Both ions start at the same point with the same velocity direction and are deflected to the same side, so the two circles are tangent at the starting point: they touch each other (IV true, III false).

Teacher's Note:
1. Radius is inversely proportional to the charge for equal mass and speed.
2. Same starting point and same velocity direction make the circles tangent to each other.

Answer: (a) I, II and IV only

Explanation:
1. Magnetic force does no work. Work-energy from P to Q: \( qE(2a)=\tfrac12m(2v)^2-\tfrac12mv^2=\tfrac32mv^2 \), so \( E=\frac{3mv^2}{4qa} \) (I true).
2. At P, the velocity is along E: \( P=qEv=q\cdot\frac{3mv^2}{4qa}\cdot v=\frac{3mv^3}{4a} \) (II true, III false).
3. At Q the velocity is along -y, perpendicular to E, and the magnetic force never does work, so the rate is zero (IV true).

Teacher's Note:
1. Use the work-energy theorem between P and Q; only the electric field does work (displacement along x is 2a).
2. Power = force dot velocity.

Answer: (b) I, II and III only

Explanation:
1. (I) No force: straight line. (II) E parallel to v changes the speed but not the direction of motion. (III) B parallel to v gives no magnetic force, and E along v keeps the path straight.
2. (IV) B perpendicular to v exerts a sideways force, so the proton is deflected: not possible.

Teacher's Note:
1. 'Undeflected' refers to direction; a force along v does not deflect.
2. A magnetic field perpendicular to v always turns the path.

Answer: (a) 2qV

Explanation:
1. In one revolution the proton crosses the gap twice, each time gaining qV.
2. Total gain per revolution = 2qV.

Teacher's Note:
1. Two gap crossings per revolution.
2. The magnetic field does no work.

Answer: (a) \( 2\pi\ \text{tesla} \)

Explanation:
1. \( f=\frac{qB}{2\pi m}\Rightarrow B=\frac{2\pi f}{q/m}=\frac{2\pi\times10^6}{10^6}=2\pi\ \text{T} \).

Teacher's Note:
1. Use the matching condition f = qB/2 pi m.
2. Keep f in hertz, not in angular units.