A charged particle carrying charge 1 μ C is moving with velocity (2i + 3j + 4k) ms–1. If an external magnetic field of (5i + 3j - 6k) × 10–3 T exists in the region where the particle is moving then the force on the particle is F × 10–9 N. The vector F is: (a) -0.30i + 0.32j - 0.09k (b) -300i + 320j - 90k (c) -30i + 32j - 9k (d) -3.0i + 3.2j - 0.9k

Answer: (c). F = q( v× B) with v×(5 i + 3 j - 6 k) for v = 2 i + 3 j + 4 k v× B = (3(-6) - 4·3) i + (4·5 - 2(-6)) j + (2·3 - 3·5) k = -30 i + 32 j - 9 k F = 10⁻⁶×10⁻³(-30 i + 32 j - 9 k) = (-30 i + 32 j - 9 k)×10⁻⁹ N So the answer is (c).

A charged particle carrying charge 1 μ C is moving with velocity (2i… | JEE Main 2020 (3 Sep, Shift 1)

Q74JEE Main 2020 · 3 SEP · SHIFT 1Medium

Moving Charge in a Magnetic Field

A charged particle carrying charge 1 $\mu$ C is moving

with velocity $\left(2\hat{i} + 3\hat{j} + 4\hat{k}\right)$ ms–1. If an external

magnetic field of $\left(5\hat{i} + 3\hat{j} - 6\hat{k}\right)$ × 10–3 T exists in the region where the particle is moving then the force on the particle is $\vec{F}$ × 10–9 N. The vector $\vec{F}$ is:

Show answer and solution

Answer: (c)

$\vec F = q(\vec v\times\vec B)$ with $\vec v\times(5\hat i + 3\hat j - 6\hat k)$ for $\vec v = 2\hat i + 3\hat j + 4\hat k$
$\vec v\times\vec B = (3(-6) - 4\cdot3)\hat i + (4\cdot5 - 2(-6))\hat j + (2\cdot3 - 3\cdot5)\hat k = -30\hat i + 32\hat j - 9\hat k$
$\vec F = 10^{-6}\times10^{-3}(-30\hat i + 32\hat j - 9\hat k) = (-30\hat i + 32\hat j - 9\hat k)\times10^{-9}\text{ N}$
So the answer is (c).

Report a mistake

Moving Charge in a Magnetic Field question from JEE Main 2020 (3 Sep, Shift 1)

More questions on Moving Charge in a Magnetic Field