The magnitude of the magnetic force on a current-carrying wire held in a magnetic

Physics

1 answer:

Kamila [148]3 years ago

3 0

Answer:

All of the above

Explanation:

The magnitude of the magnetic force on a current-carrying wire held in a magnetic is given by the equation $F = BIlsin \theta$

Where B = Strength of the magnetic field

I = The current carried by the wire

l = length of the wire in the magnetic field

θ = Angle between the wire and the magnetic field

Based on the relationship written above, the magnitude of the magnetic force on the current - carrying wire in the magnetic field depends on the strength of the magnetic field (B), length of the wire(l), current in the wire (I).

All the options are correct.

You might be interested in

As electric current moves through a wire, heat generated by resistance is conducted through a layer of insulation and then conve

mestny [16]

Answer:

Hello your question is incomplete attached below is the complete question

answer : ri ( thickness of wire ) = 14.167 m

Explanation:

attached below is a detailed solution

using the given data to determine the thickness of wire

5 0

2 years ago

Suppose the coefficient of static friction between a quarter and the back wall of a rocket car is 0.383. At what minimum rate wo

djverab [1.8K]

Answer:

25.59 m/s²

Explanation:

Using the formula for the force of static friction:

$f_s = \mu_s N$ --- (1)

where;

$f_s =$ static friction force

$\mu_s =$ coefficient of static friction

N = normal force

Also, recall that:

F = mass × acceleration

Similarly, N = mg

here, due to min. acceleration of the car;

$N = ma_{min}$

From equation (1)

$f_s = \mu_s ma_{min}$

However, there is a need to balance the frictional force by using the force due to the car's acceleration between the quarter and the wall of the rocket.

Thus,

$F = f_s$