A particle carrying a charge of 1. 0 x 10^-8C enters a magnetic field at 3.0x 10^-2 ms ^-1 at right angles to the field. If the force on this particle is 1.8 x 10^-8N, what is the magnitude of the field?

6.0x10^-1T

6.0x10T

6.0x10^-3T

6.0x10^-4T

Correct Option

4 months ago

@obedinhoagu, the answer is b, it dosent need to be changed look well, MR

10 years ago

The selected answer is wrong:

1.8/3=0.6

10'(-8)/10'(-8)*10'(-2)=10'2

0.6*10'2=60=6*10 option b

10 years ago

Please myschool.com.ng should change it to the correct option

10 years ago

my answer is B

10 years ago

Correction has been made. Thanks for your contributions.

10 years ago

The ans is B

B=F/qv=(1.8x10^-8)/(1.0x10^-8x3x10^-2)=60T or 6.0x10T

ceoofwahala

20th June, 2026

Chemistry

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ASSAAS

20th June, 2026

English Language

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infinitehoaxx

21st May, 2026

Computer

4 comments

A particle carrying a charge of 1. 0 x 10-8C enters a magnetic field at 3.0x 10-2 ms -1 at right angles to the field. If the force on this particle is 1.8 x 10-8N, what is the magnitude of the field?