A proton enters a magnetic field of flux density 1.5 Wb/m2 with a speed of 2 x 10^7 m/s at angle of 30o with the field. The force on the proton will be:
Correct answer: B. 2.4 x 10^-12N
- A. 0.24 x 10^-12N
- B. 2.4 x 10^-12N
- C. 24 x 10^-12N
- D. 0.024 x 10^-12N
Explanation
F=qvBsinθ , θ=30oVelocity of proton v=2×207m/sMagnetic field B=1.5Wb/m2Magnetic force acting on the proton F=qvBsinθ∴ F=(1.6×10−19)(2×107)(1.5)×0.5⟹ F=2.4×10−12N
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About Electromagnetism
Magnetic flux density and magnetic flux describe the strength of a magnetic field and the field passing through a surface. A charged particle moving through a magnetic field experiences a force perpendicular to its velocity and may follow circular or helical motion, depending on the angle between velocity and field.
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