CHAPTER FIVE: MAGNETIC FIELD Calculate the magnetic field strength exerting a force of 9x10-19N on a charge of 1.6x10-19C travelling at a speed of 3x106m/s perpendicular to the field?

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Asked on June 13, 2026Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

Accepted Answer · 2026-06-13T07:42:32.065Z

Here are the solutions to your magnetic field questions: 1. Calculate the magnetic field strength exerting a force of 9 × 10^-19 N on a charge of 1.6 × 10^-19 C travelling at a speed of 3 × 10^6 m/s perpendicular to the field. The magnetic force (F) on a charge (q) moving with velocity (v) in a magnetic field (B) is given by F = qvB. Since the motion is perpendicular, = 90^, so = 1. Step 1: Write down the formula and given values. F = qvB F = 9 × 10^-19 N q = 1.6 × 10^-19 C v = 3 × 10^6 m/s Step 2: Rearrange the formula to solve for B and substitute the values. B = (F)/(qv) B = 9 × 10^-19 N(1.6 × 10^-19 C)(3 × 10^6 m/s) Step 3: Calculate the magnetic field strength. B = 9 × 10^-194.8 × 10^-13 T B = 1.875 × 10^-6 T The magnetic field strength is 1.875 × 10^-6 T. 2. Calculate the magnitude of the force due to a magnetic field of 0.40 T on an electron moving perpendicular to the field with a speed of 4.00 × 10^6 m/s. (e = 1.60 × 10^-19 C) The magnetic force (F) on a charge (q) moving with velocity (v) in a magnetic field (B) is given by F = qvB. Since the motion is perpendicular, = 90^, so = 1. Step 1: Write down the formula and given values. F = qvB q = e = 1.60 × 10^-19 C v = 4.00 × 10^6 m/s B = 0.40 T Step 2: Substitute the values into the formula. F = (1.60 × 10^-19 C)(4.00 × 10^6 m/s)(0.40 T) Step 3: Calculate the magnitude of the force. F = 2.56 × 10^-13 N The magnitude of the force is 2.56 × 10^-13 N. 3. With a defined diagram explain magnetic field for both internal and external. A magnetic field is a region around a magnet or a current-carrying conductor where magnetic forces can be detected. It is represented by magnetic field lines which originate from the North pole and end at the South pole outside the magnet, and run from South to North inside the magnet, forming continuous closed loops. External Magnetic Field: This is the field observed outside a magnet or a current-carrying object. For a bar magnet, field lines emerge from the North pole and curve around to enter the South pole. For a straight current-carrying wire, the field lines form concentric circles around the wire. Internal Magnetic Field: This is the field present within the material of a magnet or inside a current-carrying coil (like a solenoid). Inside a bar magnet, the field lines run from the South pole to the North pole, completing the loop. Inside a solenoid, the field lines are nearly uniform and parallel to the axis of the coil. ` N -------------------- S | | | <---------------- | (Internal field lines S to N) | | | ----------------> | (External field lines N to S) | | ---------------------- ` (Imagine the lines curving outside from N to S and straight inside from S to N) 4. A long straight wire carries a current of 35 Amp. What is the magnitude of the field B at a point 20 cm from the wire? The magnetic field (B) at a distance (r) from a long straight wire carrying current (I) is given by B = (_0 I)/(2 r). Step 1: Convert the distance to meters and write down the given values. r = 20 cm = 0.20 m I = 35 A _0 = 4 × 10^-7 T · mA Step 2: Substitute the values into the formula. B = (4 × 10^-7 T · mA) (35 A)2 (0.20 m) Step 3: Calculate the magnetic field magnitude. B = 140 × 10^-70.40 T B = 140 × 10^-70.40 T B = 3.5 × 10^-5 T The magnitude of the field B is 3.5 × 10^-5 T. 5. A horizontal overhead power line carries a current of 90 Amp in east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line? The magnitude of the magnetic field (B) is given by B = (_0 I)/(2 r). The direction is found using the Right-Hand Rule. Step 1: Write down the given values. I = 90 A r = 1.5 m _0 = 4 × 10^-7 T · mA Step 2: Substitute the values into the formula to calculate the magnitude. B = (4 × 10^-7 T · mA) (90 A)2 (1.5 m) B = 360 × 10^-73 T B = 120 × 10^-7 T B = 1.2 × 10^-5 T Step 3: Determine the direction using the Right-Hand Rule. Point your right thumb in the direction of the current (East to West). Curl your fingers. Below the wire, your fingers will point towards the North. The magnitude of the magnetic field is 1.2 × 10^-5 T and its direction is North. 6. A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30^ with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil? The torque () on a current loop in a magnetic field is given by = NIAB. Step 1: Convert the side length to meters and calculate the area. s = 10 cm = 0.10 m A = s^2 = (0.10 m)^2 = 0.01 m^2 Step 2: Write down the given values. N = 20 turns I = 12 A B = 0.80 T = 30^ Step 3: Substitute the values into the torque formula. = (20)(12 A)(0.01 m^2)(0.80 T)(30^) = (20)(12)(0.01)(0.80)(0.5) N · m Step 4: Calculate the magnitude of the torque. = 0.96 N · m The magnitude of torque experienced by the coil is 0.96 N · m. 7. Two concentric circular coils X and Y radii 16 cm and 10 cm, respectively, lie in the same vertical plane containing the north to south direction. Coil X has 20 turns and carries a current of 16 A; coil Y has 25 turns and carries a current of 18 A. The sense of the current... The question is cut off, but it most likely asks for the net magnetic field at the center of the coils. I will assume the currents in both coils are in the same direction (e.g., both clockwise or both counter-clockwise) so their magnetic fields add up. If they were in opposite directions, the fields would subtract. The magnetic field (B) at the center of a circular coil is given by B = (_0 N I)/(2r). Step 1: Calculate the magnetic field due to Coil X. N_X = 20 r_X = 16 cm = 0.16 m I_X = 16 A B_X = (4 × 10^-7 T · mA) (20) (16 A)2 (0.16 m) B_X = 1280 × 10^-70.32 T B_X = 4000 × 10^-7 T = 4 × 10^-4 T Step 2: Calculate the magnetic field due to Coil Y. N_Y = 25 r_Y = 10 cm = 0.10 m I_Y = 18 A B_Y = (4 × 10^-7 T · mA) (25) (18 A)2 (0.10 m) B_Y = 1800 × 10^-70.20 T B_Y = 9000 × 10^-7 T = 9 × 10^-4 T Step 3: Calculate the net magnetic field at the center (assuming currents are in the same direction). B_net = B_X + B_Y B_net = (4 × 10^-4 T) + (9 × 10^-4 T) B_net = 13 × 10^-4 T B_net ≈ 13 × 3.14159 × 10^-4 T B_net ≈ 4.08 × 10^-3 T The net magnetic field at the center (assuming currents are in the same direction) is 4.08 × 10^-3 T. That's 2 down. 3 left today — send the next one.

Accepted Answer · 2026-06-13T07:42:32.065Z

Here are the solutions to your magnetic field questions: 1. Calculate the magnetic field strength exerting a force of 9 × 10^-19 N on a charge of 1.6 × 10^-19 C travelling at a speed of 3 × 10^6 m/s perpendicular to the field. The magnetic force (F) on a charge (q) moving with velocity (v) in a magnetic field (B) is given by F = qvB. Since the motion is perpendicular, = 90^, so = 1. Step 1: Write down the formula and given values. F = qvB F = 9 × 10^-19 N q = 1.6 × 10^-19 C v = 3 × 10^6 m/s Step 2: Rearrange the formula to solve for B and substitute the values. B = (F)/(qv) B = 9 × 10^-19 N(1.6 × 10^-19 C)(3 × 10^6 m/s) Step 3: Calculate the magnetic field strength. B = 9 × 10^-194.8 × 10^-13 T B = 1.875 × 10^-6 T The magnetic field strength is 1.875 × 10^-6 T. 2. Calculate the magnitude of the force due to a magnetic field of 0.40 T on an electron moving perpendicular to the field with a speed of 4.00 × 10^6 m/s. (e = 1.60 × 10^-19 C) The magnetic force (F) on a charge (q) moving with velocity (v) in a magnetic field (B) is given by F = qvB. Since the motion is perpendicular, = 90^, so = 1. Step 1: Write down the formula and given values. F = qvB q = e = 1.60 × 10^-19 C v = 4.00 × 10^6 m/s B = 0.40 T Step 2: Substitute the values into the formula. F = (1.60 × 10^-19 C)(4.00 × 10^6 m/s)(0.40 T) Step 3: Calculate the magnitude of the force. F = 2.56 × 10^-13 N The magnitude of the force is 2.56 × 10^-13 N. 3. With a defined diagram explain magnetic field for both internal and external. A magnetic field is a region around a magnet or a current-carrying conductor where magnetic forces can be detected. It is represented by magnetic field lines which originate from the North pole and end at the South pole outside the magnet, and run from South to North inside the magnet, forming continuous closed loops. External Magnetic Field: This is the field observed outside a magnet or a current-carrying object. For a bar magnet, field lines emerge from the North pole and curve around to enter the South pole. For a straight current-carrying wire, the field lines form concentric circles around the wire. Internal Magnetic Field: This is the field present within the material of a magnet or inside a current-carrying coil (like a solenoid). Inside a bar magnet, the field lines run from the South pole to the North pole, completing the loop. Inside a solenoid, the field lines are nearly uniform and parallel to the axis of the coil. ` N -------------------- S | | | <---------------- | (Internal field lines S to N) | | | ----------------> | (External field lines N to S) | | ---------------------- ` (Imagine the lines curving outside from N to S and straight inside from S to N) 4. A long straight wire carries a current of 35 Amp. What is the magnitude of the field B at a point 20 cm from the wire? The magnetic field (B) at a distance (r) from a long straight wire carrying current (I) is given by B = (_0 I)/(2 r). Step 1: Convert the distance to meters and write down the given values. r = 20 cm = 0.20 m I = 35 A _0 = 4 × 10^-7 T · mA Step 2: Substitute the values into the formula. B = (4 × 10^-7 T · mA) (35 A)2 (0.20 m) Step 3: Calculate the magnetic field magnitude. B = 140 × 10^-70.40 T B = 140 × 10^-70.40 T B = 3.5 × 10^-5 T The magnitude of the field B is 3.5 × 10^-5 T. 5. A horizontal overhead power line carries a current of 90 Amp in east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line? The magnitude of the magnetic field (B) is given by B = (_0 I)/(2 r). The direction is found using the Right-Hand Rule. Step 1: Write down the given values. I = 90 A r = 1.5 m _0 = 4 × 10^-7 T · mA Step 2: Substitute the values into the formula to calculate the magnitude. B = (4 × 10^-7 T · mA) (90 A)2 (1.5 m) B = 360 × 10^-73 T B = 120 × 10^-7 T B = 1.2 × 10^-5 T Step 3: Determine the direction using the Right-Hand Rule. Point your right thumb in the direction of the current (East to West). Curl your fingers. Below the wire, your fingers will point towards the North. The magnitude of the magnetic field is 1.2 × 10^-5 T and its direction is North. 6. A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30^ with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil? The torque () on a current loop in a magnetic field is given by = NIAB. Step 1: Convert the side length to meters and calculate the area. s = 10 cm = 0.10 m A = s^2 = (0.10 m)^2 = 0.01 m^2 Step 2: Write down the given values. N = 20 turns I = 12 A B = 0.80 T = 30^ Step 3: Substitute the values into the torque formula. = (20)(12 A)(0.01 m^2)(0.80 T)(30^) = (20)(12)(0.01)(0.80)(0.5) N · m Step 4: Calculate the magnitude of the torque. = 0.96 N · m The magnitude of torque experienced by the coil is 0.96 N · m. 7. Two concentric circular coils X and Y radii 16 cm and 10 cm, respectively, lie in the same vertical plane containing the north to south direction. Coil X has 20 turns and carries a current of 16 A; coil Y has 25 turns and carries a current of 18 A. The sense of the current... The question is cut off, but it most likely asks for the net magnetic field at the center of the coils. I will assume the currents in both coils are in the same direction (e.g., both clockwise or both counter-clockwise) so their magnetic fields add up. If they were in opposite directions, the fields would subtract. The magnetic field (B) at the center of a circular coil is given by B = (_0 N I)/(2r). Step 1: Calculate the magnetic field due to Coil X. N_X = 20 r_X = 16 cm = 0.16 m I_X = 16 A B_X = (4 × 10^-7 T · mA) (20) (16 A)2 (0.16 m) B_X = 1280 × 10^-70.32 T B_X = 4000 × 10^-7 T = 4 × 10^-4 T Step 2: Calculate the magnetic field due to Coil Y. N_Y = 25 r_Y = 10 cm = 0.10 m I_Y = 18 A B_Y = (4 × 10^-7 T · mA) (25) (18 A)2 (0.10 m) B_Y = 1800 × 10^-70.20 T B_Y = 9000 × 10^-7 T = 9 × 10^-4 T Step 3: Calculate the net magnetic field at the center (assuming currents are in the same direction). B_net = B_X + B_Y B_net = (4 × 10^-4 T) + (9 × 10^-4 T) B_net = 13 × 10^-4 T B_net ≈ 13 × 3.14159 × 10^-4 T B_net ≈ 4.08 × 10^-3 T The net magnetic field at the center (assuming currents are in the same direction) is 4.08 × 10^-3 T. That's 2 down. 3 left today — send the next one.

CHAPTER FIVE: MAGNETIC FIELD Calculate the magnetic field strength exerting a force of 9x10-19N on a charge of 1.6x10-19C travelling at a speed of 3x106m/s perpendicular to the field?

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CHAPTER FIVE: MAGNETIC FIELD Calculate the magnetic field strength exerting a force of 9x10-19N on a charge of 1.6x10-19C travelling at a speed of 3x106m/s perpendicular to the field?

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