Question:

A rectangular loop of wire with dimensions 5 cm x 10 cm is placed in a uniform magnetic field of magnitude 0.6 T. The loop is oriented such that its longer side is parallel to the direction of the magnetic field. The loop is connected to a resistor of 2 Ω as shown in the diagram below.

(a) Calculate the magnetic flux through the loop if the angle between the magnetic field and the normal to the loop is 45 degrees.

(b) Determine the emf induced in the loop when the magnetic flux changes from the value calculated in part (a) to zero in a time period of 0.5 seconds.

(c) Calculate the current that flows through the resistor when the emf is induced.

(d) How much work is done by the magnetic force when the emf changes from the value calculated in part (b) to zero?

(e) If a force is applied to compress the loop into a square shape with half the area while keeping the same magnetic field magnitude and orientation, what will be the new magnetic flux through the loop?

(f) Determine the new emf induced if the magnetic flux changes from the value calculated in part (e) to zero in a time period of 0.25 seconds.

(g) Calculate the current that flows through the resistor when the new emf is induced.

(h) Compare the power dissipated in the resistor in part (c) and part (g). Explain any differences observed.

Answer:

(a) Calculating Magnetic Flux:

The magnetic flux through a loop of area A in a uniform magnetic field B at an angle θ is given by the equation:

ϕ = B * A * cos(θ)

Given:

• B = 0.6 T (magnetic field magnitude)
• A = 5 cm * 10 cm = 50 cm²

Converting A to square meters:

• A = 50 cm² * (1 m / 100 cm)² = 0.005 m²

Substituting the values into the formula: ϕ = (0.6 T) * (0.005 m²) * cos(45°)

Using the value of cos(45°) ≈ 0.7071: ϕ = (0.6 T) * (0.005 m²) * 0.7071

Calculating ϕ: ϕ ≈ 0.0021 T·m²

Therefore, the magnetic flux through the loop is approximately 0.0021 T·m².

(b) Calculating Induced emf:

The induced emf (ε) in a loop of wire is given by the equation:

ε = - dϕ / dt

As given, the rate of change of magnetic flux (dϕ) is equal to (0.0021 T·m² - 0 T·m²) since the magnetic flux changes from the value calculated in part (a) to zero.

Substituting this value and the given time period (dt = 0.5 s) into the equation: ε = - (0.0021 T·m² - 0 T·m²) / 0.5 s

Simplifying the equation: ε = - 0.0042 V / s

Therefore, the induced emf in the loop is -0.0042 V/s.

(c) Calculating Current:

According to Ohm's Law, the current (I) flowing through a resistor (R) is given by the equation:

I = ε / R

Substituting the given values: I = (-0.0042 V/s) / (2 Ω)

Calculating I: I = -0.0021 A

Therefore, the current that flows through the resistor is approximately -0.0021 A.

(d) Calculating Work Done by Magnetic Force:

The work done by a magnetic force can be calculated using the equation:

W = -∫ ε * dq

Since the emf (ε) is constant over time, the equation simplifies to:

W = -ε * ∫ dq

The integral represents the charge (q) that flows through the resistor. To find the charge, we need to calculate the current (I) multiplied by the time (t = 0.5 s):

q = I * t

Substituting the given values: q = (-0.0021 A) * (0.5 s)

Calculating q: q = -0.00105 C

Now substituting the values into the work equation: W = -(-0.0042 V/s) * (-0.00105 C)

Simplifying W: W ≈ 0.0044 J

Therefore, the work done by the magnetic force is approximately 0.0044 Joules.

(e) Calculating New Magnetic Flux:

As given, the loop is compressed into a square shape with half the area. Therefore, the new area (A') can be calculated as follows:

A' = (1/2) * A

Substituting the given area (A = 0.005 m²): A' = (1/2) * 0.005 m²

Calculating A': A' = 0.0025 m²

Since the orientation and magnetic field magnitude remain the same, the new magnetic flux (ϕ') can be calculated using the same formula as in part (a):

ϕ' = B * A' * cos(θ)

Substituting the given values: ϕ' = (0.6 T) * (0.0025 m²) * cos(45°)

Using the value of cos(45°) ≈ 0.7071: ϕ' = (0.6 T) * (0.0025 m²) * 0.7071

Calculating ϕ': ϕ' ≈ 0.0021 T·m²

Therefore, the new magnetic flux through the loop is approximately 0.0021 T·m².

(f) Calculating New Induced emf:

Similar to part (b), the rate of change of magnetic flux (dϕ') is equal to (0.0021 T·m² - 0 T·m²) since the magnetic flux changes from the value calculated in part (e) to zero. The given time period is dt = 0.25 s.

Substituting these values into the equation: ε' = - (0.0021 T·m² - 0 T·m²) / 0.25 s

Simplifying the equation: ε' = - 0.0084 V / s

Therefore, the new induced emf in the loop is -0.0084 V/s.

(g) Calculating New Current:

Using Ohm's Law as in part (c): I' = ε' / R

Substituting the given values: I' = (-0.0084 V/s) / (2 Ω)

Calculating I': I' = -0.0042 A

Therefore, the current that flows through the resistor when the new emf is induced is approximately -0.0042 A.

(h) Comparing Power Dissipated in Resistor:

Power (P) is given by the equation:

P = I² * R

Substituting the values from part (c): P = (-0.0021 A)² * (2 Ω)

Calculating P: P ≈ 0.00000882 W

Now, substituting the values from part (g): P' = (-0.0042 A)² * (2 Ω)

Calculating P': P' ≈ 0.000035 W

The power dissipated in the resistor in part (c) is approximately 0.00000882 W, while in part (g) it is approximately 0.000035 W.

The difference in power observed is due to the change in current magnitude. In part (g), the current is twice the magnitude of the current in part (c) due to the change in magnetic flux and hence, a higher induced emf. As power is directly proportional to the square of the current, doubling the current increases the power dissipated by a factor of 4.