A proton is moving in a field of induction -2 i -4 j - T with velocity -3 - 107 i - ms -1- The force experienced by it will be-A- 0B- 6 - 10-7-k- N- 12 - 10-7-k- ND- 16-8 - 10-12-k- N

A proton enters a magnetic field with a velocity of 1.5 x 10⁵m/s making an angle of 30^owith the field. What is the force experienced by the proton if B = 3.0 T.

Q. A proton is travelling along the x-direction with velocity

5 \times 10^{6} m s^{- 1}

. The magnitude of force experienced by the proton in a magnetic field

B = (0.2^i + 0.4^k)

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A proton is moving in a field of induction [2→i+4→j]T with velocity [3×107→i]ms−1. The force experienced by it will be?

The correct option is A 16.8×10−12(+^k)N→F=q(→V×→B)=1.6×10−19[3×107^i]×[2^i+4^j]=1.6×10−19×3×107×4^k→F=16.8×10−12^k

A proton is moving in a field of induction $[2 \to i + 4 \to j] T$ with velocity $[3 \times 10^{7} \to i] m s^{- 1}$ . The force experienced by it will be?

The correct option is A $16.8 \times 10^{- 12} (+^k) N$
$\to F = q (\to V \times \to B) = 1.6 \times 10^{- 19} [3 \times 10^{7}^i] \times [2^i + 4^j] = 1.6 \times 10^{- 19} \times 3 \times 10^{7} \times 4^k \to F = 16.8 \times 10^{- 12}^k$