Questions: Moving Charges and Magnetism

Q1. State the Lorentz magnetic force on a moving charge. Why does it do no work?

Answer (English): A charge q with velocity v in a magnetic field B feels q(v × B). That vector is at right angles to both v and B, so the force never points along the motion. Work needs a force component along the displacement, and a perpendicular force has none. The path bends, while speed and kinetic energy stay fixed.

Hinglish: Moving charge par magnetic force q(v × B) hota hai, aur yeh v aur B dono ke perpendicular hota hai, isliye displacement ke bhi perpendicular hota hai. Kaam tab hota hai jab force displacement ke along ho, isliye yahan kaam zero hai. Iska matlab magnetic field direction badalta hai par speed aur kinetic energy nahi.

Q2. Find the radius of the circular orbit of a charge in a uniform field, and show the time period does not depend on the speed.

Answer (English): For motion perpendicular to a uniform field, the magnetic force q v B is constant in size and always perpendicular to the velocity, so it acts as the centripetal force. Set q v B = m v^2 / r and solve for r = m v / q B. The angular speed is v / r = q B / m, and the time period is 2 pi / angular speed = 2 pi m / q B. The speed cancels out of the time period, so a faster particle moves on a larger circle and completes it in the same time.

Hinglish: Perpendicular motion mein magnetic force q v B constant size ka hota hai aur velocity ke perpendicular hota hai, isliye woh centripetal force ka kaam karta hai. q v B = m v^2 / r se r = m v / q B. Angular speed q B / m, isliye period 2 pi m / q B. Speed cancel ho jati hai, isliye tez particle bada circle banata hai par same time mein poora karta hai.

Q3. State the Biot-Savart law and use it to obtain the magnetic field of a long straight wire.

Answer (English): The field of a current element I dl at a point with position vector r is (mu0 / 4 pi) I dl × r-hat / r^2. For a long straight wire, take a point at perpendicular distance r. Each element contributes a field perpendicular to the plane containing the element and the point, and all these contributions point the same way around the wire. Integrating along the wire from minus infinity to plus infinity gives B = mu0 I / (2 pi r). The field lines are circles centred on the wire, with direction given by the right hand grip rule.

Hinglish: Biot-Savart law kehta hai current element I dl ka field (mu nought / 4 pi) (I dl cross r hat) / r^2 hota hai. Lamba wire lo aur perpendicular distance r par point lo; sabhi elements ka field same direction mein add hota hai. Wire ke along minus infinity se plus infinity tak integrate karo to B = mu nought I / (2 pi r) milta hai. Field lines wire ke centred circles hote hain, direction right hand grip rule se.

Q4. State Ampere's law and use it to find the field inside a long solenoid.

Answer (English): Ampere's law says the line integral of B around a closed loop equals mu0 times the current passing through the loop. For a long solenoid take a rectangular loop with one long side of length L inside the solenoid parallel to the axis and the opposite side far outside. The field is uniform and parallel to the axis inside, negligible outside, and perpendicular to the short sides, so the integral reduces to B L. Each turn carries I and there are n L turns in the loop, so the enclosed current totals n L I. Ampere's law then gives B = mu0 n I inside the solenoid.

Hinglish: Ampere's law kehta hai B dot dl ka closed line integral mu nought times loop se guzarne wala current hota hai. Long solenoid ke liye rectangular loop lo: ek lamba side andar axis ke parallel, doosra bahar. Andar field uniform aur parallel hai, bahar negligible, chhote sides ke perpendicular, isliye integral B L reh jata hai. Andar enclosed current n L I, isliye B = mu nought n I milta hai.

Q5. Two long parallel wires carry currents in opposite directions. Explain the nature of the force between them.

Answer (English): Wire 1 produces a field at the position of wire 2, with size mu0 I1 / (2 pi d) where d is the separation. Wire 2 carries current I2 and feels a force per unit length of mu0 I1 I2 / (2 pi d). The direction follows from the force on a current element, I l × B. For antiparallel currents the force pushes the wires apart, so the force is repulsive. Parallel currents attract. The magnitude is the same for both wires, consistent with Newton's third law, and the force is magnetic rather than electrostatic.

Hinglish: Wire 1 apni field wire 2 ki jagah par banata hai, mu nought I1 / (2 pi d). Wire 2 par force per length mu nought I1 I2 / (2 pi d) hota hai. Direction I l cross B se nikalta hai. Antiparallel currents mein force wires ko door dhakelta hai, matlab repulsive; parallel currents attract karte hain. Magnitude dono wires par same hota hai, jo Newton ke third law se match karta hai.

Q6. Define magnetic dipole moment of a current loop and derive the torque on it in a uniform magnetic field.

Answer (English): A current loop of N turns, area A and current I has magnetic moment m = N I A, directed along the normal to the loop by the right hand rule. In a uniform field the forces on the loop segments form a couple. For a rectangular loop with its normal at angle theta to the field, the forces on the two sides perpendicular to the field are I l B each and their perpendicular separation is l cos theta, giving torque = I A B sin theta = m B sin theta. In vector form the torque is m × B, and it tends to align m with B.

Hinglish: Current loop ka magnetic moment m = N I A hota hai, direction normal ke along right hand rule se. Uniform field mein loop ke segments par forces couple banati hain. Rectangle ke liye torque I A B sin theta milta hai, matlab m B sin theta, aur vector form m cross B hai. Torque loop ke moment ko field ke along align karta hai.

Q7. Explain the working of a moving coil galvanometer and why the magnetic field is made radial.

Answer (English): A coil of N turns and area A in a radial field B carries current I and feels a torque N I A B. The suspension spring gives a restoring torque k phi. At equilibrium N I A B = k phi, so the deflection phi is proportional to the current. The radial field is made by concave pole pieces and a soft iron core so the field is always in the plane of the coil as it turns. Then the torque does not vary with deflection and the scale is linear; without it the torque would vary as the cosine of the deflection.

Hinglish: Coil par magnetic torque N I A B lagta hai aur spring restoring torque k phi deti hai. Equilibrium par N I A B = k phi, isliye deflection current ke proportional hota hai. Radial field concave poles aur soft iron core se banati hai, jisse field coil ke plane mein hi rehta hai. Tab torque deflection par depend nahi karta aur scale linear hota hai; warna torque cosine ke saath badalta.

Q8. A galvanometer of resistance G shows full scale deflection for current Ig. Derive the shunt needed to convert it into an ammeter of range I.

Answer (English): Connect a shunt resistance S in parallel with the galvanometer. The current I splits: Ig flows through the galvanometer and I minus Ig through the shunt. Both are in parallel, so the potential difference across them is equal: Ig G = (I - Ig) S. Solving gives S = Ig G / (I - Ig). The shunt is small because I minus Ig is much larger than Ig, so most of the current bypasses the coil. The equivalent resistance of the ammeter is the parallel combination, which is smaller than either resistance.

Hinglish: Galvanometer ke parallel mein shunt S lagao. Current I do hisson mein bantta hai: Ig galvanometer se aur I minus Ig shunt se. Parallel hone ki wajah se I_g G = (I - I_g) S, isliye S = I_g G / (I - I_g). Shunt chhota hota hai kyunki zyada current shunt se nikal jata hai. Ammeter ka equivalent resistance parallel combination hota hai, jo dono se chhota hota hai.

Q9. A galvanometer of resistance G and full scale current Ig is converted into a voltmeter of range V. Find the resistance required in series.

Answer (English): Connect a resistance R in series with the galvanometer so that the full scale current Ig flows when the voltage across the combination is V. The total resistance is G + R and the current is Ig, so V = Ig (G + R). Rearranging gives R = V / Ig - G. The series resistance is large, which makes the voltmeter resistance large and its current draw small. For example, with G = 20 ohm, Ig = 5 mA and V = 12 V, R = 12 / 0.005 - 20 = 2380 ohm.

Hinglish: Galvanometer ke series mein resistance R lagao taki V par full scale current Ig bache. Total resistance G + R, isliye V = I_g (G + R) aur R = V / I_g - G. Series resistance bada hota hai, isliye voltmeter ka resistance bada aur current draw chhota hota hai. Example: G = 20 ohm, Ig = 5 mA, V = 12 V par R = 12 / 0.005 - 20 = 2380 ohm.

Q10. Why does a charge moving along the axis of a long solenoid feel no magnetic force, while a charge moving at an angle follows a helix?

Answer (English): Inside a long solenoid the magnetic field is parallel to the axis. A charge moving along the axis has velocity parallel to the field, so the cross product v × B is zero and the magnetic force is zero. The charge continues in a straight line at constant speed. A charge moving at an angle to the axis has a component of velocity perpendicular to the field, and that component produces a force of size q v-perp B, which gives circular motion around the axis. The parallel component of velocity stays unchanged, so the two motions combine into a helix.

Hinglish: Long solenoid ke andar field axis ke parallel hoti hai. Axis ke along chalne wale charge ki velocity field ke parallel hoti hai, isliye v cross B zero aur force zero. Woh seedha chalta rehta hai. Angle par chalne wale charge ki velocity ka perpendicular component force deta hai q v_perp B, jo axis ke around circular motion karata hai, aur parallel component same rehta hai, isliye path helix ban jata hai.

Quiz below: five questions on the magnetic force, circular motion, Biot-Savart law, torque on a loop and galvanometer conversion. Revise the formula sheet and the galvanometer page before you start.