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. zero
Given
Charge=$$1.6*10^{-19}$$
v=(2i+3j) m/s
B=(2i+3j) wb/m
Solution
Force=q(vxB)
Cross product of v and B is 0
Therefore F is 0
The correct option is D

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