A point charge -q0 is projected in a magnetic field B-2-3k- If acceleration of the particle is a-2-b-k- then value of b will be-

The correct option is B 12 We know that the dot product of B⃗ and a⃗ will be 0 only when magnetic field exists.Thus B⃗.a⃗ =0 2+2b 3=0 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Understanding Electrostatic Potential

A point charg...

Question

A point charge $+ q_{0}$ is projected in a magnetic field $\to B = (^i + 2^j - 3^k)$ . If acceleration of the particle is $\to a = (2^i + b^j +^k)$ then value of b will be?

1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1}{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{3}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

Suggest Corrections

Similar questions

\to a = 3^i - 2^j +^k,

\to b = 2^i - 4^j - 3^k, \to c = -^i + 2^j + 2^k

\to a + \to b + \to c = 4^i - 4^j

\to a =^i -^j + 2^k, \to b = 2^i + 3^j +^k, \to c =^i -^k

\to a + 2 \to b - 3 \to c

\sqrt{78}

\to a = 2^i +^j + 3^k

\to b = 3^i + 2^j +^k

\to c =^i -^j - 4^k

\to d =^i + 2^j -^k

(\to a \times \to b) \times (\to c \times \to d) =

\to a =^i + 2^j + 3^k

\to b =^i + 2^j -^k

\to a = (1, x, - 2)

\to b = (x, 3, - 4)

x = 2

\to a =^i + 2^j + 3^k, \to b = -^i + 2^j +^k, \to c = 3^i +^j

\to a + t \to b

\to c

t = 5

\to a = 2^i + 2^j + 3^k, \to b = -^i + 2^j +^k

\to c = 3^i +^j

\to a + λ \to b

\to c

λ

Join BYJU'S Learning Program