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Electricity and Magnetism Level 2

A 3 C 3C charge moving with a velocity of 10 m / s 10m/s enters a magnetic field of strength 5 T 5 T . If its velocity makes and angle of 30 o { 30 }^{ o } with the field , then the Lorentz force experienced by the charge is

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50N 75N 25N 60N

Ankit Vijay by Ankit Vijay , 22, India

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4 solutions

Chew-Seong Cheong
Jul 29, 2014

The force F F between a current I I and magnetic field B B at an angle θ \theta is given by: F = B I s i n θ F=BIsin\theta And the current I I due to a moving charge q q at a velocity v v is given by: I = q v = 3 × 10 = 30 A I=qv=3 \times 10 = 30 A Therefore, F = B q v s i n θ = 5 × 30 × s i n 3 0 ∘ = 5 × 30 × 1 2 = 75 N F=Bqvsin\theta = 5 \times 30 \times sin30^\circ = 5 \times 30 \times \frac{1}{2} = \boxed {75 N}

12 Helpful 0 Interesting 0 Brilliant 0 Confused

but i can't see why this question asks for the magnetic field, it i devoted by E that can be accomplished by the main equation F = q.B.v sin (30 degrees),

then from this result, we insert it at such equation like this E=F/q thus, the final result we get is: E=25N/C

Pedró Bahy - 6 years, 9 months ago
Samarpit Swain
Jul 28, 2014

Force experienced by a moving charge ( Q ) (Q) in a uniform magnetic field ( B ) (B) with velocity ( V ) (V) is given by: F = Q V B s i n ( A ) . . . . . . . ( i ) [ H e r e ′ A ′ = 30 d e g ] F=QVBsin(A) .......(i) [Here 'A'= 30 deg]

Plugging the values in equation(i) gives: F = 75 N F= 75 N

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Raymond Lin
Aug 4, 2014

The magnetic force acting on a charge is described by the equation F = q ( v × B ) F=q(v \times B) , so the magnitude of the force is q v B sin ⁡ θ qvB\sin\theta . Plugging in the given values, we get that the answer is 3 ∗ 10 ∗ 5 ∗ sin ⁡ 3 0 ∘ 3*10*5*\sin{30^\circ} , or 75 \fbox{75} .

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F=Bqvsin(theta)

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