The force F on a current-carrying wire in a magnetic field is given by:
F=ILBsinθ
where:
- I is the current (3 A),
- L is the length of the wire (0.5 m),
- B is the magnetic field strength (0.4 T),
- θ is the angle between the wire and the magnetic field (here, θ=90∘).
Since sin90∘=1:
F=ILB
Substitute the given values:
F=3×0.5×0.4
F=0.6 N
So, the force on the wire is 0.6 N.
Problem 4: Magnetic Field at the Center of a Circular Loop
Problem: A circular loop of radius 0.1 m carries a current of 2 A. Calculate the magnetic field strength at the center of the loop.
Solution:
The magnetic field B at the center of a circular loop is given by:
B=2Rμ0I
where:
- μ0 is the permeability of free space (4π×10−7 T⋅m/A),
- I is the current (2 A),
- R is the radius of the loop (0.1 m).
Substitute the given values:
B=2×0.14π×10−7×2
B=0.28π×10−7
B=0.14π×10−6
B=4π×10−5
Using π≈3.14:
B≈4×3.14×10−5
B≈1.256×10−4 T
So, the magnetic field strength at the center of the loop is approximately 1.256×10−4 T.
Problem 5: Induced EMF in a Moving Conductor
Problem: A straight conductor of length 0.2 m moves with a velocity of 5 m/s perpendicular to a magnetic field of 0.3 T. Calculate the induced EMF in the conductor.
Solution:
The induced EMF E in a moving conductor is given by:
E=BvL
where:
- B is the magnetic field strength (0.3 T),
- v is the velocity of the conductor (5 m/s),
- L is the length of the conductor (0.2 m).
Substitute the given values:
E=0.3×5×0.2
E=0.3×1
E=0.3 V
So, the induced EMF in the conductor is 0.3 V.
Problem 6: Magnetic Force on a Charged Particle in a Uniform Magnetic Field
Problem: An electron (charge q=1.6×10−19 C) moves in a uniform magnetic field of 0.1 T with a velocity of 1×107 m/s making an angle of 30∘ with the magnetic field. Calculate the magnitude of the magnetic force on the electron.
Solution:
The magnetic force F on a charged particle moving at an angle θ with the magnetic field is given by:
F=qvBsinθ
where:
- q is the charge of the electron (1.6×10−19 C),
- v is the velocity (1×107 m/s),
- B is the magnetic field strength (0.1 T),
- θ is the angle between the velocity and the magnetic field (30∘).
Since sin30∘=0.5:
F=qvBsin30∘
Substitute the given value
F
F=0.6 N
So, the force on the wire is 0.6 N.