A proton enters a magnetic field with a velocity of 2.5×107ms−1 making an angle 30∘ with the magnetic field. The force on the proton is (B=25T)

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A proton enters a magnetic field with a velocity of {2.5}{10}^

Question

A proton enters a magnetic field with a velocity of $2.5 \times 10^{7} m s^{- 1}$ making an angle $30^{\circ}$ with the magnetic field. The force on the proton is $(B = 25 T)$

1.25 \times 10^{- 11} N

2.5 \times 10^{- 11} N

5.0 \times 10^{- 11} N

7.5 \times 10^{- 11} N

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A proton enters a magnetic field with a velocity of

2.5 \times 10^{7} m s^{- 1}

making an angle

30^{\circ}

with the magnetic field. The force on the proton is

(B = 25 T)

8 mins ago

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Question Text	A proton enters a magnetic field with a velocity of $2.5 \times 10^{7} m s^{- 1}$ making an angle $30^{\circ}$ with the magnetic field. The force on the proton is $(B = 25 T)$
Answer Type	Video solution: 1
Upvotes	151