Q125 / 200 · NEET 2021 · Physics

In the product

F→=q(v→×B→)

=qv→×Bi^+Bj^+B0k^

For q=1 and v→=2i^+4j^+6k^ and F→=4i^-20j^+12k^

What will be the complete expression for B→?

  • A
    -8ı^-8j^-6k^
  • Correct answer
    B
    -6ı^-6j^-8k^
  • C
    8i^+8j^-6k^
  • D
    6i^+6j^-8k^

Solution

4ı^-20ȷ^+12k=(2ı^+4ȷ^+6k^)×Bı^+Bȷ^+Bok^

4ı^-20ȷ^+12k^=+ı^-ȷ^+k^246BBB0

=4B0-6Bı^-2B0-6Bȷ^+(2B-4B)k^

4B0-6B=4

2B0-6B=20

2B-4B=12⇒B=6

So, 2B0-6(-6)=20

2B0=20-36⇒B0=-8

∴B→=-6ı^-6ȷ^-8k^

Hence, the correct answer is option: 'B'.

Relevant Questions