Question

# A test $$1.6 \times 10^{-19} Cb$$ is moving with velocity $$\overrightarrow v = ( 2 \hat i +3 \hat j)$$ m/sec is magnetic field $$\overrightarrow B = ( 2 \hat i + 3 \hat j ) wb /m^2$$ .the magnetic force on the test charge:-

A
$$6 \hat k T$$
B
$$( 4 \hat i + 6 \hat j ) T$$
C
$$( 4 \hat i + 6 \hat j ) \times 10^{-19} T$$
D
zero
Solution
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#### Correct option is D. zeroGivenCharge=$$1.6*10^{-19}$$v=(2i+3j) m/sB=(2i+3j) wb/mSolutionForce=q(vxB)Cross product of v and B is 0Therefore F is 0The correct option is D

0
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