Inside a solenoid: field, force on a wire, and an electron's circle

SPH4U Grade 12 Physics · Gravitational, Electric and Magnetic Fields

A demonstration solenoid is wound with 1200 turns over a length of 40.0 cm and carries 3.50 A. A probe wire 5.00 cm long is clamped across the bore at right angles to the axis and run at 8.00 A so the force on it can be read on a balance. The probe is then removed and an electron is injected across the bore at 2.40 × 10⁶ m/s, again at right angles to the field. Find the field inside the coil, the force on the probe wire, the magnetic force on the electron, and the radius of the circle the electron travels.

Step 1 of 4 · solve for Magnetic field

Everything in this chain is downstream of B. Note what the formula actually depends on: turns per metre, not turns — a longer coil with the same 1200 turns gives a weaker field. The coil is 40.0 cm long, so the 0.400 m in the denominator is the conversion doing its job.

Rearranged for B
B=μ0NILB = \tfrac{\mu_0 N I}{L}
Your values, in your units
B=μ0(1,200)(3.5 A)(40 cm)B = \tfrac{\mu_0 \, \left( 1{,}200 \right) \, \left( 3.5\ \text{A} \right)}{\left( 40\ \text{cm} \right)}
Converted to base units
B=μ0(1,200)(3.5 A)(0.4 m)B = \tfrac{\mu_0 \, \left( 1{,}200 \right) \, \left( 3.5\ \text{A} \right)}{\left( 0.4\ \text{m} \right)}
Answer
B=13.2B = 13.2

Carried onward at full precision, not this rounded figure.

Open the Magnetic Field of a Solenoid solver →

Step 2 of 4 · solve for Magnetic force

A current balance is how you check a field you only calculated. Only the 5.00 cm inside the bore counts, and only because it sits at 90° to the field — the leads running parallel to B feel nothing.

Rearranged for F
F=BILsinθF = B I L \sin\theta
13.2 mTcarried from step 1
Your values, in your units
F=(0.0131947 T)(8 A)(5 cm)sin(90 )F = \left( 0.0131947\ \text{T} \right) \, \left( 8\ \text{A} \right) \, \left( 5\ \text{cm} \right) \, \sin \left( 90\ ^{\circ} \right)
Converted to base units
F=(0.0131947 T)(8 A)(0.05 m)sin(90 )F = \left( 0.0131947\ \text{T} \right) \, \left( 8\ \text{A} \right) \, \left( 0.05\ \text{m} \right) \, \sin \left( 90\ ^{\circ} \right)
Answer
F=528F = 528

Carried onward at full precision, not this rounded figure.

Open the Magnetic Force on a Current-Carrying Wire solver →

Step 3 of 4 · solve for Magnetic force

Same field, different carrier of the current: one electron instead of the 5 × 10¹⁹ per second the probe wire was pushing. F = qvB sin θ is the microscopic law that F = BIL sin θ is built out of.

Rearranged for F
F=qvBsinθF = q v B \sin\theta
13.2 mTcarried from step 1
Your values, in your units
F=(1 e)(2,400,000 m/s)(0.0131947 T)sin(90 )F = \left( 1\ \text{e} \right) \, \left( 2{,}400{,}000\ \text{m/s} \right) \, \left( 0.0131947\ \text{T} \right) \, \sin \left( 90\ ^{\circ} \right)
Converted to base units
F=(1.60218e19 C)(2,400,000 m/s)(0.0131947 T)sin(90 )F = \left( 1.60218e-19\ \text{C} \right) \, \left( 2{,}400{,}000\ \text{m/s} \right) \, \left( 0.0131947\ \text{T} \right) \, \sin \left( 90\ ^{\circ} \right)
Answer
F=5.07e15F = 5.07e-15

Carried onward at full precision, not this rounded figure.

Open the Magnetic Force on a Moving Charge solver →

Step 4 of 4 · solve for Radius

The magnetic force is always perpendicular to the velocity, so it never changes the speed — it can only bend the path. Set it equal to mv²/r and the radius of that bend falls out.

Rearranged for r
r=mv2Fcr = \frac{m v^2}{F_c}
5.07e-15 Ncarried from step 3
Your values, in your units
r=(9.10938e31 kg)(2,400,000 m/s)2(5.07365e15 N)r = \frac{\left( 9.10938e-31\ \text{kg} \right) \, \left( 2{,}400{,}000\ \text{m/s} \right)^2}{\left( 5.07365e-15\ \text{N} \right)}
Answer
r=1.03r = 1.03

Carried onward at full precision, not this rounded figure.

Open the Centripetal Force (F = mv²/r) solver →

Why this order

Read this chain as one field interrogated three ways. Step 1 predicts B from the winding; step 2 puts a current-carrying wire in it and produces a force a balance can read, which is how you find out whether step 1 was honest; steps 3 and 4 replace the wire with a single electron. The order cannot be shuffled, because B is the only quantity all three share and both force laws are useless until it exists. The classic error in step 1 is reading N as turns per metre: 1200 turns over 40.0 cm is 3000 turns per metre, and dropping the division altogether leaves the field 2.5 times too small. The length is also where the centimetres go — divide by 40 instead of by 0.400 m and the field comes out a hundred times too small, which is a mistake no amount of care with μ₀ will catch.

Step 4 carries the real idea. A magnetic force on a moving charge is always at right angles to the motion, so it does no work at all — the electron leaves the field with exactly the 2.40 × 10⁶ m/s it came in with, just pointed somewhere else. That is what makes the path a circle rather than a curve that speeds up or slows down. Setting qvB = mv²/r gives r = mv/qB, so the radius depends on momentum per unit charge and on nothing else, which is why the same apparatus in a school lab measures the electron's charge-to-mass ratio and, scaled up, why the Large Hadron Collider needs 8 tesla dipoles to hold protons on a 27 km ring. Watch the sin θ as well: the 90° here makes it 1 and quietly disappears, but an electron injected along the axis instead of across it would feel no force whatever and sail straight through.

Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.