# 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:-

**B**

$$ ( 4 \hat i + 6 \hat j ) T $$

**C**

$$ ( 4 \hat i + 6 \hat j ) \times 10^{-19} T $$

#### 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**