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Chapter 4: 55E. (page 916)

You wish to hit a target from several meters away with a charged coin having a mass of \(4.25\,{\rm{g}}\) and a charge of \( + 2500\,{\rm{\mu C}}\). The coin is given an initial velocity of

12.8m/s,and a downward, uniform electric field with field strength 27.5 N/Cexists throughout the region. If you aim directly at the target and fire the coin horizontally, what magnitude and direction of uniform magnetic field are needed in the region for the coin to hit the target?

Short Answer

Expert verified

The magnitude and direction of uniform magnetic field are needed in the region for the coin to hit the target are \(3.45\,{\rm{T}}\)and to the left when viewed on the direction of the target.

Step by step solution

01

Magnetic and electrical force

The magnetic force and electrical force

\(qvB = qE + mg\)

02

Step 2: Identification of the given data

The value of mass is,\(m = 4.25\,{\rm{g}}\)

The value of charge is,\(q = 2500\,{\rm{\mu C}}\).

The initial velocity is, v = 112.8m/s

The value of field strength is, E = 27.5N/C.

03

Determine the magnitude of magnetic force and direction

The magnetic force is given by

\(qvB = qE + mg\)

Therefore,

\(B = \frac{{qE + mg}}{{qv}}\)

Substitute all the value in the above equation.

\(\begin{aligned}{}B = \frac{{\left( {2500 \times {{10}^{ - 6}}\,{\rm{C}}} \right)(27.5\,{\rm{N/C}}) + (0.00425\,{\rm{kg}})\left( {9.80\,{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)}}{{\left( {2500 \times {{10}^{ - 6}}\,{\rm{C}}} \right)(12.8\,{\rm{m/s}})}}\\ = 3.45\;\,{\rm{T}}\end{aligned}\)

Since the direction of the electric field is downwards, so the magnetic force is upwards.

Therefore, the direction of the magnetic field is into the page.

Therefore, the magnetic field is \(3.45\;\,{\rm{T}}\)

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