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Question6. A conductor 25 cm long carries a current of 50.0 A . It is placed in a magnetic field of strength 49 T . Determine the force exerted on the conductor when it makes each of the following angles with the magnetic lines of force ( 3 marks) a) 0∘0^{\circ} b) 45∘45^{\circ} c) 90∘90^{\circ}

Studdy Solution

STEP 1

What is this asking? Find the force on a wire in a magnetic field at different angles. Watch out! Don't forget to convert the length to meters and use the angle in radians for calculations!

STEP 2

1. Understand the formula
2. Convert units
3. Calculate the force for each angle

STEP 3

Alright, let's start by understanding the formula we need!
The force on a current-carrying conductor in a magnetic field is given by:
F=I⋅L⋅B⋅sin⁡(θ)F = I \cdot L \cdot B \cdot \sin(\theta)where: - F F is the force in newtons (N), - I I is the current in amperes (A), - L L is the length of the conductor in meters (m), - B B is the magnetic field strength in teslas (T), - θ \theta is the angle between the conductor and the magnetic field in radians.
This formula tells us that the force depends on the sine of the angle between the wire and the magnetic field.
So, when the angle is 0∘0^\circ, the sine is zero, and when it's 90∘90^\circ, the sine is one!

STEP 4

Next up, let's convert the length of the conductor from centimeters to meters because our formula needs it in meters.
We have:
25 cm=0.25 m25 \, \text{cm} = 0.25 \, \text{m}

STEP 5

Let's start with part (a), where the angle is 0∘0^\circ.
The sine of 0∘0^\circ is zero, so:
F=50.0⋅0.25⋅49⋅sin⁡(0)=0 NF = 50.0 \cdot 0.25 \cdot 49 \cdot \sin(0) = 0 \, \text{N}The force is zero because the conductor is parallel to the magnetic field lines!

STEP 6

Now for part (b), where the angle is 45∘45^\circ.
We need to use the sine of 45∘45^\circ, which is 22\frac{\sqrt{2}}{2}:
F=50.0⋅0.25⋅49⋅sin⁡(π4)=50.0⋅0.25⋅49⋅22F = 50.0 \cdot 0.25 \cdot 49 \cdot \sin\left(\frac{\pi}{4}\right) = 50.0 \cdot 0.25 \cdot 49 \cdot \frac{\sqrt{2}}{2}Calculating this gives us:
F≈433.01 NF \approx 433.01 \, \text{N}

STEP 7

Finally, part (c), where the angle is 90∘90^\circ.
The sine of 90∘90^\circ is one, so:
F=50.0⋅0.25⋅49⋅sin⁡(π2)=50.0⋅0.25⋅49⋅1F = 50.0 \cdot 0.25 \cdot 49 \cdot \sin\left(\frac{\pi}{2}\right) = 50.0 \cdot 0.25 \cdot 49 \cdot 1This simplifies to:
F=612.5 NF = 612.5 \, \text{N}

STEP 8

The forces exerted on the conductor at different angles are: - At 0∘0^\circ: 0 N0 \, \text{N} - At 45∘45^\circ: approximately 433.01 N433.01 \, \text{N} - At 90∘90^\circ: 612.5 N612.5 \, \text{N}

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