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2 years ago

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A charge of 8.4 × 10–4 C moves at an angle of 35° to a magnetic field that has a field strength of 6.7 × 10–3 T. If the magnetic

force is 3.5 × 10–2 N, how fast is the charge moving?

1 answer:

larisa86 [58]2 years ago

4 0

Answer:

The charge is moving with the velocity of $1.1\times10^{4}\ m/s$ .

Explanation:

Given that,

Charge $q =8.4\times10^{-4}\ C$

Angle = 35°

Magnetic field strength $B=6.7\times10^{-3}\ T$

Magnetic force $F=3.5\times10^{-2}\ N$

We need to calculate the velocity.

The Lorentz force exerted by the magnetic field on a moving charge.

The magnetic force is defined as:

$F = qvB\sin\theta$

$v = \dfrac{F}{qB\sin\theta}$

Where,

F = Magnetic force

q = charge

B = Magnetic field strength

v = velocity

Put the value into the formula

$v =\dfrac{3.5\times10^{-2}}{8.4\times10^{-4}\times6.7\times10^{-3}\times\sin35^{\circ}}$

$v =\dfrac{3.5\times10^{-2}}{8.4\times10^{-4}\times6.7\times10^{-3}\times0.57}$

$v = 10910.36\ m/s$

$v = 1.1\times10^{4}\ m/s$

Hence, The charge is moving with the velocity of $1.1\times10^{4}\ m/s$ .

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