A proton moving with a velocity of 6- - 8- - 105 ms-1 enters uniform magnetic field of induction 5 - 10-3k-T - The magnitude of the force acting on the proton is - - - and k- are unit vectors forming a right handed triad

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 moving with a velocity of

3 \times 10^{7} m / s

in the direction of a uniform magnetic field of

0.5 T

. The force acting on proton is

Q. A beam of protons enters a uniform magnetic field of

0.3 T

with a velocity of

4 \times 10^{5} m / s

in a direction making an angle of

60^{\circ}

with the direction of magnetic field. What will be the pitch of the helix formed by moving particles? (Given charge of proton

e = 1.6 \times 10^{- 19} C

, mass of proton

m = 1.67 \times 10^{- 27} k g

)

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A proton moving with a velocity of (6^i+8^j)×105 ms−1 enters uniform magnetic field of induction 5×10−3^kT . The magnitude of the force acting on the proton is :(^i,^j and ^k are unit vectors forming a right handed triad)

The correct option is B 8×10−16NForce on a charged particle in magnetic field is given by,¯F=q(→V×→B)=1.6×10−19((6^i+8^j)×5^k)×105×10−3=1.6×10−19×10−2(−30^j+40^i)=1.6×10−16(−3^j+4^i)|F|=1.6×10−16√32+42=1.6×10−16×5N=8×10−16N

A proton moving with a velocity of $(6^i + 8^j) \times 10^{5} m s^{- 1}$ enters uniform magnetic field of induction $5 \times 10^{- 3}^k T$ . The magnitude of the force acting on the proton is :
( $^i,^j$ and $^k$ are unit vectors forming a right handed triad)