PHYSICS
JAMB 2012 - Question 41
Physics 2012 JAMB Past Questions - Question 41: A particle carrying a change of 1.0x10-2¸C enters a magnetic field at 3.0x10 ms-¹ at right angles to the field .If the force on this particle is 1.8x10-2N. What is the magnitude of the field ?
Choose the correct answers from the options given.
A:
B:
C:
D:
Correct Answer
C
Explanation
To find the magnitude of the magnetic field, we can use the formula for the magnetic force on a charged particle moving through a magnetic field:
F = q * v * B
Where:
F is the force on the particle (given as 1.8x10^(-8) N),
q is the charge of the particle (given as 1.0x10^(-8) C),
v is the velocity of the particle (given as 3.0x10^5 m/s),
B is the magnitude of the magnetic field (what we need to find).
Rearranging the formula, we can solve for B:
B = F / (q * v)
Substituting the given values:
B = (1.8x10^(-8) N) / (1.0x10^(-8) C * 3.0x10^5 m/s)
Calculating this expression:
B = 6.0x10^(-5) T
Therefore, the magnitude of the magnetic field is 6.0x10^(-5) Tesla.
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