A conductor of length 0.8 m carrying a current of 2.5 A is kept in a magnetic field of 0.6 Tesla. What is the magnitude of the force experienced by the conductor if it is inclined at an angle 30° to the direction of the magnetic field?
Choose one answer.
a. 0.6 N
b. 1.0 N
c. 1.2 N
d. 1.8 N
a. 0.6 N
F = B I L sin theta
To find the magnitude of the force experienced by the conductor, we can use the formula for the magnetic force on a current-carrying conductor:
F = BILsinθ
Where:
F is the magnitude of the force,
B is the magnetic field strength,
I is the current in the conductor,
L is the length of the conductor, and
θ is the angle between the direction of the current and the magnetic field.
In this case, we are given:
B = 0.6 Tesla (magnetic field strength),
I = 2.5 A (current),
L = 0.8 m (length of the conductor), and
θ = 30° (angle between the conductor and the magnetic field).
Plugging in these values into the formula, we can calculate the force:
F = (0.6 Tesla) * (2.5 A) * (0.8 m) * sin(30°)
sin(30°) = 0.5 (value of sine for 30°)
F = (0.6 Tesla) * (2.5 A) * (0.8 m) * 0.5
F = 0.6 N
Therefore, the magnitude of the force experienced by the conductor when inclined at an angle of 30° to the magnetic field is 0.6 Newtons.
So, the correct answer is (a) 0.6 N.