A charged particle moving with a velocity \overrightarrow{v}=v_1 \stackrel{\^}{i}+v_2 \stackrel{\^}{j} in a magnetic field \overrightarrow{B} experiences a force \overrightarrow{F}=F_1 \stackrel{\^}{i}+F_2 \stackrel{\^}{j}. Here v_1, v_2, F_1, F_2 all are constants. Then { {\B}}^{\to} can be

Correct Answer is : →B=B1^i+B2^j+B3^k with v1v2=B1B2

\overrightarrow{B}=B_1 \stackrel{\^}{i}+B_2 \stackrel{\^}{j} with \frac{v_1}{v_2}=\frac{B_1}{B_2}

\overrightarrow{B}=B_1 \stackrel{\^}{i}+B_2 \stackrel{\^}{j}+B_3 \stackrel{\^}{k} with \frac{v_1}{v_2}=\frac{B_1}{B_2}

{ {\B}}^{\to}= {\B}_3 \stackrel{\^}{ {\j}} with {\B}_1= {\B}_2=0

\overrightarrow{B}=B_1 \stackrel{\^}{j}+B_2 \stackrel{\^}{k} with \frac{B_1}{B_2}=\frac{v_1}{v_2}

WBJEE 2024 Physics & Chemistry Paper

Question : 66 of 80

Marks: +1, -0

A charged particle moving with a velocity

→

=v1

+v2

in a magnetic field

→

experiences a force

→

=F1

+F2

. Here v1,v2,F1,F2 all are constants. Then

→

can be

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