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 ?

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.