Dedicated Solved Examples — Full Chapter Coverage

Welcome to the practice arena of Chapter 4. The 30 problems that follow are a complete tour of everything covered in this chapter.

Recommended workflow — two passes:

  • Pass 1 (45–60 mins). Cover the printed solutions. Attempt each problem yourself within ~2 minutes. Note where you got stuck.
  • Pass 2 (15–30 mins). Read the worked solutions in full. Pay close attention to direction reasoning, unit consistency, and where formulas come from.

If you can solve 22+ of these 30 without hints, you are board-ready and JEE/NEET-ready for this chapter.

How to Get the Most Out of This Section

A few study tips that make the difference between passive reading and active learning.

  1. Time-budget — 90 seconds per problem on the first attempt. If you cannot identify which law applies in 90 seconds, mark it as a concept gap and revisit Section N before continuing.
  2. Attempt blind first. Use a fresh sheet, write down the data, draw the geometry, identify the unknown. Only then check the printed solution.
  3. Revisit the ones you got wrong after 24 hours. The neuroscience of spaced repetition says a wrong answer reviewed the next day is worth ten reviewed immediately.
  4. Don't memorise solutions — memorise the trigger word. "Wire perpendicular to B" →\to F=BILF = BIL. "Charge at angle to B" →\to helix. "Field at centre of loop" →\to μ0I/(2R)\mu_0 I/(2R). Build these reflexes.
  5. Always check units at the end. A field that comes out in N/(C⋅m)\text{N}/(\text{C}\cdot\text{m}) instead of tesla is wrong.
  6. Sanity-check direction using a right-hand rule even when only magnitude is asked — many MCQ traps swap the correct magnitude with a wrong sign.
  7. Keep a one-page "mistake log" of every error you make in these 30 problems. On exam morning, review only that log.

Solved Examples

Example 1: Lorentz force — simple magnitude

A proton moves with velocity v⃗=4×105 j^\vec{v} = 4 \times 10^{5}\,\hat{j} m/s in a magnetic field B⃗=0.3 k^\vec{B} = 0.3\,\hat{k} T. Find the magnetic force on the proton.

Solution. F⃗=qv⃗×B⃗=(1.6×10−19)(4×105 j^)×(0.3 k^)\vec{F} = q\vec{v}\times\vec{B} = (1.6\times 10^{-19})(4\times 10^{5}\,\hat{j})\times(0.3\,\hat{k}).

j^×k^=i^\hat{j}\times\hat{k} = \hat{i}, so

F⃗=(1.6×10−19)(4×105)(0.3) i^=1.92×10−14 i^ N.\vec{F} = (1.6\times 10^{-19})(4\times 10^{5})(0.3)\,\hat{i} = 1.92 \times 10^{-14}\,\hat{i}\ \text{N}.

Magnitude F=1.92×10−14 N\boxed{F = 1.92\times 10^{-14}\ \text{N}}, direction +i^+\hat{i}.

Example 2: Force on a straight wire

A 20 cm long wire carrying current 5 A is placed perpendicular to a uniform magnetic field of 0.4 T. Find the magnitude of the force on the wire.

Solution. F=BILsin⁡θF = BIL\sin\theta with θ=90°\theta = 90°.

F=(0.4)(5)(0.20)(1)=0.40 N.F = (0.4)(5)(0.20)(1) = 0.40\ \text{N}.

F=0.40 N\boxed{F = 0.40\ \text{N}}, perpendicular to both the wire and B⃗\vec{B}.

Example 3: Radius of a charged particle's circular path

An electron (mass 9.1×10−319.1\times 10^{-31} kg, charge 1.6×10−191.6\times 10^{-19} C) enters a magnetic field of 0.01 T perpendicularly with a speed of 2×1062\times 10^{6} m/s. Find the radius of its circular trajectory.

Solution. r=mvqBr = \dfrac{mv}{qB}.

r=(9.1×10−31)(2×106)(1.6×10−19)(0.01)=1.82×10−241.6×10−21≈1.14×10−3 m.r = \frac{(9.1\times 10^{-31})(2\times 10^{6})}{(1.6\times 10^{-19})(0.01)} = \frac{1.82\times 10^{-24}}{1.6\times 10^{-21}} \approx 1.14\times 10^{-3}\ \text{m}.

r≈1.14 mm\boxed{r \approx 1.14\ \text{mm}}.

Example 4: BB from a long straight wire

Find the magnetic field at a perpendicular distance of 5 cm from a long straight wire carrying a current of 10 A.

Solution. B=μ0I2πrB = \dfrac{\mu_0 I}{2\pi r}.

B=(4π×10−7)(10)2π(0.05)=4×10−60.10=4×10−5 T.B = \frac{(4\pi\times 10^{-7})(10)}{2\pi (0.05)} = \frac{4\times 10^{-6}}{0.10} = 4\times 10^{-5}\ \text{T}.

B=4×10−5 T=40 μT\boxed{B = 4\times 10^{-5}\ \text{T} = 40\ \mu\text{T}}.

Example 5: BB at the centre of a circular loop

A circular coil of radius 10 cm has 100 turns and carries a current of 2 A. Find the magnetic field at its centre.

Solution. B=μ0NI2RB = \dfrac{\mu_0 N I}{2R}.

B=(4π×10−7)(100)(2)2(0.10)=8π×10−50.20=4π×10−4≈1.26×10−3 T.B = \frac{(4\pi\times 10^{-7})(100)(2)}{2(0.10)} = \frac{8\pi\times 10^{-5}}{0.20} = 4\pi\times 10^{-4} \approx 1.26\times 10^{-3}\ \text{T}.

B≈1.26×10−3 T=1.26 mT\boxed{B \approx 1.26\times 10^{-3}\ \text{T} = 1.26\ \text{mT}}.

Example 6: Charge moving parallel to B⃗\vec{B}

A 1 C charge moves with velocity v⃗=5 i^\vec{v} = 5\,\hat{i} m/s in a magnetic field B⃗=2 i^\vec{B} = 2\,\hat{i} T. Find the magnetic force on it.

Solution. F⃗=qv⃗×B⃗\vec{F} = q\vec{v}\times\vec{B}. Here v⃗∥B⃗\vec{v}\parallel\vec{B}, so v⃗×B⃗=0\vec{v}\times\vec{B} = 0.

F⃗=0.\boxed{\vec{F} = 0.}

A magnetic field exerts no force on a charge moving parallel (or antiparallel) to it. This is why the field component along the velocity has no dynamical effect.

Example 7: Period of circular motion

A charged particle of mass m=2×10−27m = 2 \times 10^{-27} kg and charge q=1.6×10−19q = 1.6\times 10^{-19} C moves in a magnetic field of 0.5 T. Find its period of revolution.

Solution. T=2πmqBT = \dfrac{2\pi m}{qB} — independent of speed.

T=2π(2×10−27)(1.6×10−19)(0.5)=4π×10−278×10−20≈1.57×10−7 s.T = \frac{2\pi(2\times 10^{-27})}{(1.6\times 10^{-19})(0.5)} = \frac{4\pi\times 10^{-27}}{8\times 10^{-20}} \approx 1.57\times 10^{-7}\ \text{s}.

T≈1.57×10−7 s=0.157 μs\boxed{T \approx 1.57\times 10^{-7}\ \text{s} = 0.157\ \mu\text{s}}.

Example 8: Solenoid field — simple plug-in

A long solenoid has 2000 turns per metre and carries a current of 3 A. Find the magnetic field inside it.

Solution. B=μ0nIB = \mu_0 n I.

B=(4π×10−7)(2000)(3)=2.4π×10−3≈7.54×10−3 T.B = (4\pi\times 10^{-7})(2000)(3) = 2.4\pi\times 10^{-3} \approx 7.54\times 10^{-3}\ \text{T}.

B≈7.54×10−3 T=7.54 mT\boxed{B \approx 7.54\times 10^{-3}\ \text{T} = 7.54\ \text{mT}}.

Example 9: Magnetic moment of a current loop

A circular loop of radius 5 cm carries a current of 4 A. Find the magnitude of its magnetic dipole moment.

Solution. m=IA=I(πr2)m = IA = I(\pi r^2).

m=4×π(0.05)2=4π(2.5×10−3)=π×10−2≈3.14×10−2 A m2.m = 4 \times \pi (0.05)^2 = 4\pi (2.5\times 10^{-3}) = \pi \times 10^{-2} \approx 3.14\times 10^{-2}\ \text{A m}^2.

m≈3.14×10−2 A m2\boxed{m \approx 3.14\times 10^{-2}\ \text{A m}^2}.

Example 10: Force between two parallel wires

Two long parallel wires 10 cm apart carry currents of 5 A each in the same direction. Find the force per metre between them.

Solution. FL=μ0I1I22πd\dfrac{F}{L} = \dfrac{\mu_0 I_1 I_2}{2\pi d}.

FL=(4π×10−7)(5)(5)2π(0.10)=10−50.10=5×10−5 N/m.\frac{F}{L} = \frac{(4\pi\times 10^{-7})(5)(5)}{2\pi(0.10)} = \frac{10^{-5}}{0.10} = 5\times 10^{-5}\ \text{N/m}.

Same direction ⇒\Rightarrow attractive. F/L=5×10−5 N/m, attractive\boxed{F/L = 5\times 10^{-5}\ \text{N/m, attractive}}.

Example 11: Cyclotron — kinetic energy

A cyclotron accelerates protons to a maximum radius of 0.50 m using a magnetic field of 1.2 T. Find the kinetic energy of the protons in MeV. (Take mp=1.67×10−27m_p = 1.67\times 10^{-27} kg, e=1.6×10−19e = 1.6\times 10^{-19} C.)

Solution. From r=mv/(qB)r = mv/(qB), v=qBr/mv = qBr/m. Then

K=12mv2=q2B2r22m.K = \tfrac{1}{2}mv^2 = \frac{q^2 B^2 r^2}{2m}.

K=(1.6×10−19)2(1.2)2(0.5)22(1.67×10−27)=(2.56×10−38)(1.44)(0.25)3.34×10−27.K = \frac{(1.6\times 10^{-19})^2 (1.2)^2 (0.5)^2}{2(1.67\times 10^{-27})} = \frac{(2.56\times 10^{-38})(1.44)(0.25)}{3.34\times 10^{-27}}.

Numerator =9.22×10−39= 9.22\times 10^{-39}. So K=2.76×10−12 JK = 2.76\times 10^{-12}\ \text{J}.

Convert: K=2.76×10−12/(1.6×10−13)≈17.2K = 2.76\times 10^{-12}/(1.6\times 10^{-13}) \approx 17.2 MeV.

K≈17 MeV\boxed{K \approx 17\ \text{MeV}}.

Example 12: Cyclotron frequency

Find the cyclotron frequency for a deuteron (m=3.34×10−27m = 3.34\times 10^{-27} kg, q=eq = e) in a magnetic field of 0.8 T.

Solution. fc=qB2πmf_c = \dfrac{qB}{2\pi m}.

fc=(1.6×10−19)(0.8)2π(3.34×10−27)=1.28×10−192.10×10−26≈6.1×106 Hz.f_c = \frac{(1.6\times 10^{-19})(0.8)}{2\pi (3.34\times 10^{-27})} = \frac{1.28\times 10^{-19}}{2.10\times 10^{-26}} \approx 6.1\times 10^{6}\ \text{Hz}.

fc≈6.1 MHz\boxed{f_c \approx 6.1\ \text{MHz}}.

Example 13: Magnetic field of a finite straight wire

A straight wire of length 20 cm carries a current of 5 A. Find the magnetic field at a perpendicular distance of 10 cm from the centre of the wire, measured on the perpendicular bisector.

Solution. For a finite straight wire,

B=μ0I4πd(sin⁡θ1+sin⁡θ2).B = \frac{\mu_0 I}{4\pi d}(\sin\theta_1 + \sin\theta_2).

By symmetry θ1=θ2=θ\theta_1 = \theta_2 = \theta where tan⁡θ=(L/2)/d=10/10=1\tan\theta = (L/2)/d = 10/10 = 1, so θ=45°\theta = 45°, sin⁡θ=1/2\sin\theta = 1/\sqrt{2}.

B=(4π×10−7)(5)4π(0.10)⋅(2⋅12)=5×10−70.10⋅2=(5×10−6)2≈7.07×10−6 T.B = \frac{(4\pi\times 10^{-7})(5)}{4\pi (0.10)} \cdot (2 \cdot \tfrac{1}{\sqrt{2}}) = \frac{5\times 10^{-7}}{0.10}\cdot \sqrt{2} = (5\times 10^{-6})\sqrt{2} \approx 7.07\times 10^{-6}\ \text{T}.

B≈7.07 μT\boxed{B \approx 7.07\ \mu\text{T}}.

Example 14: Ampère's law inside a thick wire

A long cylindrical conductor of radius R=5R = 5 cm carries a uniformly distributed current of 10 A. Find the magnetic field at r=2r = 2 cm from the axis.

Solution. Inside (r<Rr < R), Ampère's law with enclosed current Ienc=I(r2/R2)I_{enc} = I(r^2/R^2) gives

B=μ0Ir2πR2.B = \frac{\mu_0 I r}{2\pi R^2}.

B=(4π×10−7)(10)(0.02)2π(0.05)2=(4×10−8)2.5×10−3=1.6×10−5 T.B = \frac{(4\pi\times 10^{-7})(10)(0.02)}{2\pi (0.05)^2} = \frac{(4\times 10^{-8})}{2.5\times 10^{-3}} = 1.6\times 10^{-5}\ \text{T}.

B=1.6×10−5 T=16 μT\boxed{B = 1.6\times 10^{-5}\ \text{T} = 16\ \mu\text{T}}.

Example 15: Solenoid — finding nn from total turns

A solenoid of length 50 cm has 500 turns and carries a current of 4 A. Find BB at its centre. Treat it as ideal.

Solution. n=N/L=500/0.50=1000 turns/mn = N/L = 500/0.50 = 1000\ \text{turns/m}.

B=μ0nI=(4π×10−7)(1000)(4)=1.6π×10−3≈5.03×10−3 T.B = \mu_0 n I = (4\pi\times 10^{-7})(1000)(4) = 1.6\pi\times 10^{-3} \approx 5.03\times 10^{-3}\ \text{T}.

B≈5.03 mT\boxed{B \approx 5.03\ \text{mT}}.

Example 16: Force per unit length — opposite currents

Two parallel wires 5 cm apart carry currents 8 A and 12 A in opposite directions. Find the force per metre between them.

Solution. FL=μ0I1I22πd\dfrac{F}{L} = \dfrac{\mu_0 I_1 I_2}{2\pi d}.

FL=(4π×10−7)(8)(12)2π(0.05)=(2×10−7)(96)0.05=3.84×10−4 N/m.\frac{F}{L} = \frac{(4\pi\times 10^{-7})(8)(12)}{2\pi (0.05)} = \frac{(2\times 10^{-7})(96)}{0.05} = 3.84\times 10^{-4}\ \text{N/m}.

Opposite directions ⇒\Rightarrow repulsive. F/L=3.84×10−4 N/m, repulsive\boxed{F/L = 3.84\times 10^{-4}\ \text{N/m, repulsive}}.

Example 17: Torque on a current loop

A rectangular coil of 50 turns and area 4×10−3 m24\times 10^{-3}\ \text{m}^2 carries a current of 2 A. It is placed in a magnetic field of 0.5 T, with its plane parallel to B⃗\vec{B}. Find the torque.

Solution. When the coil's plane is parallel to B⃗\vec{B}, the area vector is perpendicular to B⃗\vec{B}, so θ=90°\theta = 90° and sin⁡θ=1\sin\theta = 1.

τ=NIABsin⁡θ=(50)(2)(4×10−3)(0.5)(1)=0.20 N m.\tau = NIAB\sin\theta = (50)(2)(4\times 10^{-3})(0.5)(1) = 0.20\ \text{N m}.

τ=0.20 N m\boxed{\tau = 0.20\ \text{N m}}.

Example 18: Galvanometer — current sensitivity

A moving coil galvanometer of resistance 50 Ω\Omega shows full-scale deflection for a current of 5 mA. What is its current sensitivity if the maximum deflection is 30 divisions?

Solution. Current sensitivity =deflectioncurrent=305×10−3=6000 div/A=6 div/mA= \dfrac{\text{deflection}}{\text{current}} = \dfrac{30}{5\times 10^{-3}} = 6000\ \text{div/A} = 6\ \text{div/mA}.

SI=6 div/mA\boxed{S_I = 6\ \text{div/mA}}.

Example 19: Helical motion

A proton enters a magnetic field of 0.10 T at an angle of 60° to the field with a speed of 2×1062\times 10^6 m/s. Find the pitch of the helical path.

Solution. v∥=vcos⁡60°=106v_\parallel = v\cos 60° = 10^6 m/s. Period T=2πm/(qB)T = 2\pi m/(qB).

T=2π(1.67×10−27)(1.6×10−19)(0.10)=1.05×10−261.6×10−20≈6.56×10−7 s.T = \frac{2\pi (1.67\times 10^{-27})}{(1.6\times 10^{-19})(0.10)} = \frac{1.05\times 10^{-26}}{1.6\times 10^{-20}} \approx 6.56\times 10^{-7}\ \text{s}.

Pitch p=v∥T=(106)(6.56×10−7)≈0.66p = v_\parallel T = (10^6)(6.56\times 10^{-7}) \approx 0.66 m.

p≈0.66 m\boxed{p \approx 0.66\ \text{m}}.

Example 20: Force between long wires — net force on a third wire

Three long parallel wires lie in the same plane, each carrying 10 A in the same direction. The spacing is 5 cm between adjacent wires. Find the force per metre on the middle wire.

Solution. The middle wire feels two equal-magnitude forces in opposite directions (one wire pulls left, the other pulls right). By symmetry, they cancel.

Fnet/L=0\boxed{F_{net}/L = 0} on the middle wire.

Example 21: Velocity selector (combined E⃗\vec{E} and B⃗\vec{B})

In a region, E⃗=104 j^\vec{E} = 10^4\,\hat{j} V/m and B⃗=0.02 k^\vec{B} = 0.02\,\hat{k} T. A positive charge moves along +i^+\hat{i} with speed vv. For what vv does it pass undeflected?

Solution. Net force is zero when electric and magnetic forces cancel.

qE⃗+qv⃗×B⃗=0q\vec{E} + q\vec{v}\times\vec{B} = 0. Take v⃗=vi^\vec{v} = v\hat{i}: v⃗×B⃗=v(0.02)(i^×k^)=−0.02v j^\vec{v}\times\vec{B} = v(0.02)(\hat{i}\times\hat{k}) = -0.02v\,\hat{j}.

So q(104−0.02v)j^=0⇒v=104/0.02=5×105q(10^4 - 0.02 v)\hat{j} = 0 \Rightarrow v = 10^4/0.02 = 5\times 10^{5} m/s.

v=5×105 m/s\boxed{v = 5\times 10^{5}\ \text{m/s}} — independent of charge or mass.

Example 22: Railgun — force on a sliding rod

A conducting rod of length 0.40 m and mass 50 g slides on frictionless rails carrying a current of 20 A. A uniform magnetic field of 0.30 T is perpendicular to the plane of the rails. Find the acceleration of the rod.

Solution. Force on the rod: F=BIL=(0.30)(20)(0.40)=2.4F = BIL = (0.30)(20)(0.40) = 2.4 N.

Acceleration: a=F/m=2.4/0.05=48 m/s2a = F/m = 2.4/0.05 = 48\ \text{m/s}^2.

a=48 m/s2\boxed{a = 48\ \text{m/s}^2}.

Example 23: Multi-loop superposition — concentric coils

Two concentric circular coils lie in the same plane. Coil 1 has radius 10 cm, 50 turns, current 2 A. Coil 2 has radius 20 cm, 100 turns, current 3 A, in the opposite sense. Find the net field at the common centre.

Solution. B1=μ0N1I12R1=(4π×10−7)(50)(2)2(0.10)=2π×10−4 TB_1 = \dfrac{\mu_0 N_1 I_1}{2R_1} = \dfrac{(4\pi\times 10^{-7})(50)(2)}{2(0.10)} = 2\pi\times 10^{-4}\ \text{T}.

B2=μ0N2I22R2=(4π×10−7)(100)(3)2(0.20)=3π×10−4 TB_2 = \dfrac{\mu_0 N_2 I_2}{2R_2} = \dfrac{(4\pi\times 10^{-7})(100)(3)}{2(0.20)} = 3\pi\times 10^{-4}\ \text{T}.

Opposite sense ⇒\Rightarrow Bnet=∣B2−B1∣=π×10−4≈3.14×10−4 TB_{net} = |B_2 - B_1| = \pi\times 10^{-4} \approx 3.14\times 10^{-4}\ \text{T}, in the direction of coil 2.

Bnet≈3.14×10−4 T\boxed{B_{net} \approx 3.14\times 10^{-4}\ \text{T}}.

Example 24: Converting a galvanometer to an ammeter

A galvanometer of resistance G=100 ΩG = 100\ \Omega gives full-scale deflection at Ig=1I_g = 1 mA. To convert it into an ammeter of range 0–5 A, what shunt resistance is needed?

Solution. Shunt S=IgGI−IgS = \dfrac{I_g G}{I - I_g}.

S=(10−3)(100)5−10−3≈0.105=0.020 Ω.S = \frac{(10^{-3})(100)}{5 - 10^{-3}} \approx \frac{0.10}{5} = 0.020\ \Omega.

S≈0.020 Ω=20 mΩ\boxed{S \approx 0.020\ \Omega = 20\ \text{m}\Omega}, connected in parallel with the galvanometer.

Example 25: Converting a galvanometer to a voltmeter

The same galvanometer (G=100 ΩG = 100\ \Omega, Ig=1I_g = 1 mA) is to be converted into a voltmeter of range 0–10 V. What series resistance is needed?

Solution. Series resistance R=VIg−GR = \dfrac{V}{I_g} - G.

R=1010−3−100=10000−100=9900 Ω.R = \frac{10}{10^{-3}} - 100 = 10000 - 100 = 9900\ \Omega.

R=9900 Ω=9.9 kΩ\boxed{R = 9900\ \Omega = 9.9\ \text{k}\Omega}, connected in series.

Example 26: Field at the centre of a square loop

A square loop of side length a=20a = 20 cm carries a current of 4 A. Find the magnitude of the magnetic field at its geometric centre.

Solution. The geometric centre of a square loop lies at a perpendicular distance of d=a/2=10d = a/2 = 10 cm =0.1= 0.1 m from each side. Each of the four straight wire segments subtends angles of α1=α2=45∘\alpha_1 = \alpha_2 = 45^\circ at the centre.

The magnetic field contributed by a single side is given by: B1=μ0I4πd(sin⁡α1+sin⁡α2)=μ0I4π(a/2)(sin⁡45∘+sin⁡45∘)B_1 = \frac{\mu_0 I}{4\pi d}(\sin\alpha_1 + \sin\alpha_2) = \frac{\mu_0 I}{4\pi (a/2)}(\sin 45^\circ + \sin 45^\circ)

B1=μ0I2πa(12+12)=μ0I2πa(2)=2μ0I2πaB_1 = \frac{\mu_0 I}{2\pi a} \left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right) = \frac{\mu_0 I}{2\pi a}(\sqrt{2}) = \frac{\sqrt{2}\mu_0 I}{2\pi a}

Since the current flows in a continuous loop, the right-hand rule shows that the magnetic field contributions from all four sides point in the exact same direction (into or out of the page). Thus, we multiply by 4: Btotal=4B1=4×2μ0I2πa=22μ0IπaB_{\text{total}} = 4 B_1 = 4 \times \frac{\sqrt{2}\mu_0 I}{2\pi a} = \frac{2\sqrt{2}\mu_0 I}{\pi a}

Substituting our values (I=4I = 4 A, a=0.2a = 0.2 m, and μ0=4π×10−7\mu_0 = 4\pi \times 10^{-7} T·m/A): Btotal=22×(4π×10−7)×4π×0.20=322×10−70.20=1602×10−7≈2.26×10−5 TB_{\text{total}} = \frac{2\sqrt{2} \times (4\pi \times 10^{-7}) \times 4}{\pi \times 0.20} = \frac{32\sqrt{2} \times 10^{-7}}{0.20} = 160\sqrt{2} \times 10^{-7} \approx 2.26 \times 10^{-5}\,\text{T}

Btotal≈2.26×10−5 T=22.6 μT\boxed{B_{\text{total}} \approx 2.26 \times 10^{-5}\,\text{T} = 22.6\,\mu\text{T}}

Example 27: Field at the centre of a regular hexagonal loop

A regular hexagonal loop of side length a=10a = 10 cm carries a current of 5 A. Find the magnetic field at its centre.

Solution. A regular hexagon consists of 6 equal sides. Each side forms an equilateral triangle with the geometric centre, meaning each side subtends a total angle of 60∘60^\circ. The perpendicular distance dd from the centre to any side is given by: d=a2tan⁡30∘=a32=0.10×32=0.053 md = \frac{a}{2\tan 30^\circ} = \frac{a\sqrt{3}}{2} = \frac{0.10 \times \sqrt{3}}{2} = 0.05\sqrt{3}\,\text{m}

For a single side segment, the angles measured from the perpendicular bisector to the ends are α1=α2=30∘\alpha_1 = \alpha_2 = 30^\circ. The field from one side is: B1=μ0I4πd(sin⁡30∘+sin⁡30∘)=μ0I4πd(12+12)=μ0I4πdB_1 = \frac{\mu_0 I}{4\pi d}(\sin 30^\circ + \sin 30^\circ) = \frac{\mu_0 I}{4\pi d}\left(\frac{1}{2} + \frac{1}{2}\right) = \frac{\mu_0 I}{4\pi d}

By symmetry, all 6 sides contribute fields in the same direction. The total magnetic field is: Btotal=6B1=6μ0I4πd=3μ0I2π(a32)=3μ0IπaB_{\text{total}} = 6 B_1 = \frac{6\mu_0 I}{4\pi d} = \frac{3\mu_0 I}{2\pi \left(\frac{a\sqrt{3}}{2}\right)} = \frac{\sqrt{3}\mu_0 I}{\pi a}

Substituting the numerical values (I=5I = 5 A, a=0.10a = 0.10 m, and μ0=4π×10−7\mu_0 = 4\pi \times 10^{-7} T·m/A): Btotal=3×(4π×10−7)×5π×0.10=203×10−70.10=2003×10−7≈3.46×10−5 TB_{\text{total}} = \frac{\sqrt{3} \times (4\pi \times 10^{-7}) \times 5}{\pi \times 0.10} = \frac{20\sqrt{3} \times 10^{-7}}{0.10} = 200\sqrt{3} \times 10^{-7} \approx 3.46 \times 10^{-5}\,\text{T}

Btotal≈3.46×10−5 T=34.6 μT\boxed{B_{\text{total}} \approx 3.46 \times 10^{-5}\,\text{T} = 34.6\,\mu\text{T}}

Example 28: Combined E⃗\vec{E} and B⃗\vec{B} — charge gains energy

A proton starts from rest, is accelerated through a potential difference of 1 kV, then enters a magnetic field of 0.10 T perpendicular to its velocity. Find the radius of its circular trajectory.

Solution. Speed from the potential drop: 12mv2=qV⇒v=2qV/m\tfrac{1}{2}m v^2 = qV \Rightarrow v = \sqrt{2qV/m}.

v=2(1.6×10−19)(1000)1.67×10−27=1.92×1011≈4.38×105 m/s.v = \sqrt{\frac{2(1.6\times 10^{-19})(1000)}{1.67\times 10^{-27}}} = \sqrt{1.92\times 10^{11}} \approx 4.38\times 10^{5}\ \text{m/s}.

Radius: r=mv/(qB)=(1.67×10−27)(4.38×105)/[(1.6×10−19)(0.10)]r = mv/(qB) = (1.67\times 10^{-27})(4.38\times 10^{5})/[(1.6\times 10^{-19})(0.10)].

r=(7.31×10−22)/(1.6×10−20)≈4.57×10−2 mr = (7.31\times 10^{-22})/(1.6\times 10^{-20}) \approx 4.57\times 10^{-2}\ \text{m}.

r≈4.6 cm\boxed{r \approx 4.6\ \text{cm}}.

Example 29: Torque and potential energy — combined

A circular coil (50 turns, radius 5 cm, current 2 A) is placed in B⃗=0.20\vec{B} = 0.20 T. Initially the magnetic moment is parallel to B⃗\vec{B}. The coil is rotated through 180°. Find (a) the maximum torque during rotation, (b) the work done by the external agent.

Solution.

Magnetic moment: m=NIA=50×2×π(0.05)2=0.785 A m2m = NIA = 50 \times 2 \times \pi (0.05)^2 = 0.785\ \text{A m}^2.

(a) Max torque occurs at θ=90°\theta = 90°: τmax=mB=0.785×0.20=0.157 N m\tau_{max} = mB = 0.785 \times 0.20 = 0.157\ \text{N m}.

(b) Energy: U=−mBcos⁡θU = -mB\cos\theta. Initial Ui=−mBU_i = -mB; final Uf=+mBU_f = +mB. Work done by external agent against the field:

W=Uf−Ui=2mB=2(0.785)(0.20)=0.314 J.W = U_f - U_i = 2mB = 2(0.785)(0.20) = 0.314\ \text{J}.

τmax=0.157 N m, W=0.314 J\boxed{\tau_{max} = 0.157\ \text{N m},\ W = 0.314\ \text{J}}.

Example 30: Mass spectrometer — separating isotopes

Two isotopes of mass m1=20um_1 = 20 u and m2=22um_2 = 22 u are singly ionised and accelerated through the same potential difference V=5000V = 5000 V, then enter a magnetic field B=0.50B = 0.50 T perpendicular to their motion. Find the difference in their radii. (1 u=1.66×10−271\,u = 1.66\times 10^{-27} kg.)

Solution. Speed: v=2qV/mv = \sqrt{2qV/m}. Radius: r=mv/(qB)=1B2mV/qr = mv/(qB) = \dfrac{1}{B}\sqrt{2mV/q}.

For m1=20×1.66×10−27=3.32×10−26m_1 = 20 \times 1.66\times 10^{-27} = 3.32\times 10^{-26} kg:

r1=10.52(3.32×10−26)(5000)/(1.6×10−19)r_1 = \dfrac{1}{0.5}\sqrt{2(3.32\times 10^{-26})(5000)/(1.6\times 10^{-19})}.

Inside the square root: 2(3.32×10−26)(5000)/(1.6×10−19)=(3.32×10−22)/(1.6×10−19)=2.075×10−32(3.32\times 10^{-26})(5000)/(1.6\times 10^{-19}) = (3.32\times 10^{-22})/(1.6\times 10^{-19}) = 2.075\times 10^{-3}.

2.075×10−3=4.56×10−2\sqrt{2.075\times 10^{-3}} = 4.56\times 10^{-2}. So r1=9.11×10−2r_1 = 9.11\times 10^{-2} m ≈9.11\approx 9.11 cm.

For m2=22 um_2 = 22\,u: r2=r122/20=9.11×1.0488≈9.56r_2 = r_1\sqrt{22/20} = 9.11 \times 1.0488 \approx 9.56 cm.

Difference Δr=r2−r1≈0.45\Delta r = r_2 - r_1 \approx 0.45 cm =4.5= 4.5 mm.

Δr≈4.5 mm\boxed{\Delta r \approx 4.5\ \text{mm}} — small but enough to physically separate isotopes onto different detectors.