Year 11 Physics Module 4 · IQ3 25 min 8 MC · 3 SA

Checkpoint 3 — Inquiry Question 3

This checkpoint assesses your understanding of magnetism and magnetic fields (Lessons 12–14): magnetic field lines, and the magnetic fields produced by current-carrying wires and solenoids.

Multiple Choice

4 × 1 = 4 MARKS

UnderstandBand 4

1. A long straight wire carries current upward. What is the direction of the magnetic field at a point directly to the east of the wire? 1 MARK

A. North

B. South

C. East

D. Upward

Select your answer above.

Write A, B, C or D in your book.

RememberBand 3

2. Which formula correctly gives the magnetic field inside a long solenoid? 1 MARK

A. $B = \mu_0 I / (2\pi r)$

B. $B = \mu_0 N I / L$

C. $B = \mu_0 I / d$

D. $B = N I / L$

ApplyBand 5

3. A long straight wire carries 10 A. What is the magnetic field at a distance of 5.0 cm from the wire? ($\mu_0 = 4\pi \times 10^{-7}\ \text{T m A}^{-1}$) 1 MARK

A. $2.0 \times 10^{-6}\ \text{T}$

B. $4.0 \times 10^{-6}\ \text{T}$

C. $4.0 \times 10^{-5}\ \text{T}$

D. $8.0 \times 10^{-5}\ \text{T}$

ApplyBand 5

4. A solenoid has 300 turns in a length of 15 cm and carries 2.0 A. What is the magnetic field inside? ($\mu_0 = 4\pi \times 10^{-7}\ \text{T m A}^{-1}$) 1 MARK

A. $1.26 \times 10^{-3}\ \text{T}$

B. $5.03 \times 10^{-3}\ \text{T}$

C. $2.51 \times 10^{-2}\ \text{T}$

D. $1.01 \times 10^{-2}\ \text{T}$

Short Answer

1 × 4 = 4 MARKS

ApplyBand 5

5. A solenoid has 500 turns wound uniformly over a length of 25 cm. It carries a current of 4.0 A. ($\mu_0 = 4\pi \times 10^{-7}\ \text{T m A}^{-1}$) (a) Calculate the magnetic field inside the solenoid. (b) Explain why the field inside a long solenoid is approximately uniform and parallel to the axis. (c) State one factor that would decrease the magnetic field strength if changed, and explain why. 4 MARKS

Answer in your book

Saved

Comprehensive Answers

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Multiple Choice

B — Right-hand grip rule: thumb up (current), fingers curl. At east, fingers point south.
B — $B = \mu_0 N I / L$ for a solenoid. A is the formula for a straight wire.
C — $B = \mu_0 I/(2\pi r) = (2 \times 10^{-7})(10)/(0.050) = 4.0 \times 10^{-5}\ \text{T}$.
B — $B = \mu_0 N I / L = (4\pi \times 10^{-7})(300)(2.0)/(0.15) = 5.03 \times 10^{-3}\ \text{T}$.

Q5 — Model Answer (4 marks)

(a) $B = \mu_0 N I / L = (4\pi \times 10^{-7})(500)(4.0)/(0.25) = (4\pi \times 10^{-7})(2000)/(0.25) = 2.51 \times 10^{-3}\ \text{T}$ (or approximately $2.5 \times 10^{-3}\ \text{T}$).

(b) Inside a long solenoid, the magnetic field lines from each turn of the coil are parallel to the axis and add together. Near the centre, the field from one turn is reinforced by the field from all neighbouring turns, producing a uniform field. At the ends, some field lines spread outward, causing slight non-uniformity, but for a long solenoid this end effect is negligible over most of the interior.

(c) Decreasing the current would decrease the field because $B \propto I$. Alternatively, increasing the solenoid length (while keeping turns constant) would decrease $n = N/L$, reducing $B$ because $B \propto n$. Another valid answer: removing the iron core (if present) would decrease $B$ because ferromagnetic materials amplify the field through domain alignment.