Question

Завдання 4. Яка сила струму у провіднику, який знаходиться в магнітному полі з індукціею 2 Тл. якщо довжнна активної частини провідника 20 см, а сила, діюча на провідник, 70 мН? Кут між напрямком лінії індукції і струмом 30%.

Answer 1

Дано:

В= 2 Тл

L= 20 см= 0,2 м

Fа= 70 мН= 7×10⁴ Н

α= 30°

_______________

І-?

Розв'язок:

$Fa = ВІL sin \alpha$

звідси знайдемо силу струму в провіднику:

$І = \frac{Fа}{ВL \sin \alpha }$

$І = \frac{7 \times {10}^{4} }{2 \times 0.2 \times0.5 } = 35 \times {10}^{4} А$

Відповідь: І= 350 кА

Answer 2

To find the current in the conductor, we can use the formula for the force on a current-carrying conductor in a magnetic field:

F = BIL sin(θ)

Where: - F is the force on the conductor, - B is the magnetic field strength (induction), - I is the current in the conductor, - L is the length of the active part of the conductor, and - θ is the angle between the direction of the magnetic field and the current.

In this case, we are given: - B = 2 T (induction), - L = 20 cm = 0.2 m (length of the active part of the conductor), - F = 70 mN = 70 * 10^(-3) N (force on the conductor), and - θ = 30% = 30° (angle between the direction of the magnetic field and the current).

To find the current I, we rearrange the formula:

I = F / (B * L * sin(θ))

Now we can substitute the given values into the formula and calculate the current.

Calculation:

I = 70 * 10^(-3) N / (2 T * 0.2 m * sin(30°))

Using a calculator, we can evaluate the expression:

I ≈ 0.7 A

Therefore, the current in the conductor is approximately 0.7 Amperes.

Please note that the above calculation assumes a linear conductor and neglects any resistance in the circuit.