Electric Fields

Reading cue. r is drawn from the wire to the near face of the probe, not to the sensor inside it, and the field being measured points perpendicular to the page, not along the line r.

1. aState State how the direction of the magnetic field at the probe is related to the plane of the drawing, and state whether the field points the same way once the probe is moved to the dashed position.

   recall3 marks

   Check answer · 3 marks

   1. Right-hand grip rule applied to the current arrow drawn on the wire
   2. Field at the probe is perpendicular to the plane of the drawing, not along the dimension line r
   3. Field at the dashed position points the same way, because that position lies on the same side of the wire; only the magnitude falls

2. bCalculate The wire carries a steady current of 4.0 A. The dimension r is 25 mm for the probe drawn in full and 100 mm for the dashed position. Calculate the magnetic flux density at each of these two positions.

   routine3 marks

   Check answer · 3 marks

   1. Use of B = μ₀I/(2πr), with r the perpendicular distance drawn from the wire
   2. At r = 0.025 m: B = (2 × 10⁻⁷ × 4.0)/0.025 = 3.2 × 10⁻⁵ T (32 μT)
   3. At r = 0.100 m: B = (2 × 10⁻⁷ × 4.0)/0.100 = 8.0 × 10⁻⁶ T (8.0 μT)

3. cDetermine The dimension r is drawn to the near face of the probe, but the sensing element sits 5.0 mm behind that face. Determine the percentage by which each recorded flux density falls below μ₀I/(2πr) when r is recorded as 25 mm and when it is recorded as 100 mm, and state which end of a graph of B against 1/r is distorted more.

   demanding4 marks

   Check answer · 4 marks

   1. True separation is r + 5.0 mm, so the recorded value is in the ratio r/(r + 5.0 mm) of the expected one
   2. At r = 25 mm: 25/30, so the reading is 17% below the expected value
   3. At r = 100 mm: 100/105, so the reading is 4.8% below the expected value
   4. Readings taken closest to the wire are affected most, so the graph bends below a straight line at the large-1/r end

4. dDiscuss The wire is drawn extending well above and well below the level of the probe. Discuss what happens to the readings at both probe positions if the wire is shortened until its ends lie only a few centimetres above and below that level.

   top of the paper4 marks

   Check answer · 4 marks

   1. μ₀I/(2πr) assumes an infinitely long wire, so that current elements at all distances along it contribute
   2. A shortened wire is missing those contributions, so every measured value of B is smaller than μ₀I/(2πr)
   3. The shortfall grows with r, since the shortened wire subtends a smaller angle at the probe, so the dashed far position is affected more than the near one
   4. The plotted line therefore falls away from a straight line at the small-1/r end, and its gradient no longer gives the true current

Reading cue. The crosses mean the field is into the page and the drawn arc bends downwards; the charge is negative, so check your rule reproduces that downward force before trusting any later step.

1. aState State the direction of the magnetic force on the electron at the instant it crosses the dashed boundary, and state the direction of the conventional current that the moving electron represents.

   recall2 marks

   Check answer · 2 marks

   1. Conventional current is directed to the left, opposite to the drawn motion of the electron
   2. Force is directed downwards, towards the bottom of the page, in agreement with the way the drawn arc bends

2. bDetermine Determine the time the electron would take to complete one full circle inside the field region, and state how that time would differ if the electron had been accelerated through 2000 V instead of 500 V.

   routine4 marks

   Check answer · 4 marks

   1. Use of T = 2πm/(qB)
   2. T = 2π × 9.11 × 10⁻³¹/(1.60 × 10⁻¹⁹ × 2.5 × 10⁻³)
   3. T = 1.4 × 10⁻⁸ s
   4. Time would be unchanged, because the period does not depend on the speed

3. cDetermine The dashed region is 5.0 cm wide in the direction of the electron's initial motion, and is tall enough that the electron never reaches its upper or lower edge. Determine the greatest distance the electron penetrates into the region, and determine where it leaves.

   demanding4 marks

   Show a hint

   The centre of the circular path lies on the perpendicular to the velocity drawn at the entry point.

   Check answer · 4 marks

   1. Speed on entry v = √(2eV/m) = 1.3 × 10⁷ m s⁻¹
   2. Radius r = mv/(eB) = 1.21 × 10⁻²³/(4.0 × 10⁻²²) = 3.0 × 10⁻² m
   3. Greatest penetration equals r, i.e. 3.0 cm after a quarter circle, which is less than the 5.0 cm width
   4. Electron turns through 180° and leaves through the boundary it entered, 2r = 6.0 cm below the entry point

4. dEvaluate The accelerating potential difference is raised from 500 V to 1000 V with everything else unchanged. Evaluate whether the electron can be made to leave through the far edge of the region drawn, either by this change or by raising the accelerating potential difference further.

   top of the paper4 marks

   Check answer · 4 marks

   1. r ∝ √V, so the radius rises by a factor of √2, from 3.0 cm to 4.3 cm
   2. Penetration still equals r, and 4.3 cm is less than the 5.0 cm width, so the electron still turns back
   3. Penetration reaches 5.0 cm when r = 5.0 cm, requiring V = 500 × (5.0/3.0)² ≈ 1.4 × 10³ V
   4. Above about 1.4 kV the electron does reach the far edge, so the behaviour drawn is a consequence of this particular accelerating voltage rather than a general feature of the arrangement