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Ch 27: Magnetic Field and Magnetic Forces
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610당신이 사용하는 게 아니라요?교과서 변경
27장, 문제 17

A 150 g ball containing 4.00 x 108 excess electrons is dropped into a 125 m vertical shaft. At the bottom of the shaft, the ball suddenly enters a uniform horizontal magnetic field that has magnitude 0.250 T and direction from east to west. If air resistance is negligibly small, find the magnitude and direction of the force that this magnetic field exerts on the ball just as it enters the field.

검증된 단계별 안내
1
First, calculate the charge of the ball using the number of excess electrons. The charge of one electron is approximately \(-1.6 \times 10^{-19}\) C. Multiply this by the number of excess electrons \(4.00 \times 10^8\) to find the total charge \(q\) on the ball.
Next, identify the velocity of the ball as it enters the magnetic field. Since the ball is dropped from a height of 125 m, use the kinematic equation \(v^2 = u^2 + 2gh\) where \(u = 0\) (initial velocity), \(g = 9.8\) m/s² (acceleration due to gravity), and \(h = 125\) m (height) to find the final velocity \(v\).
Now, apply the formula for the magnetic force \(F = qvB\sin\theta\), where \(q\) is the charge calculated in step 1, \(v\) is the velocity from step 2, \(B = 0.250\) T is the magnetic field strength, and \(\theta\) is the angle between the velocity and the magnetic field direction. Since the ball enters the field horizontally, \(\theta = 90^\circ\), making \(\sin\theta = 1\).
Calculate the magnitude of the force using the values obtained: \(F = qvB\). Substitute the values of \(q\), \(v\), and \(B\) into the equation to find the force.
Determine the direction of the force using the right-hand rule. Point your fingers in the direction of the velocity (downward), curl them towards the direction of the magnetic field (east to west), and your thumb will point in the direction of the force. This will help you identify the direction of the force exerted on the ball.

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이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.
영상 길이:
7m
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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Magnetic Force on a Moving Charge

The magnetic force on a moving charge is given by the Lorentz force equation, F = q(v × B), where q is the charge, v is the velocity, and B is the magnetic field. This force is perpendicular to both the velocity of the charge and the magnetic field direction, influencing the trajectory of charged particles in a magnetic field.
추천 영상:
가이드 코스
09:08
Magnetic Force Between Two Moving Charges

Charge Calculation from Excess Electrons

The charge of an object can be calculated from the number of excess electrons using the formula q = n × e, where n is the number of excess electrons and e is the elementary charge (approximately 1.6 × 10^-19 C). This calculation is crucial for determining the force exerted by the magnetic field on the charged ball.
추천 영상:
가이드 코스
04:43
Calculating Displacement from Velocity-Time Graphs

Gravitational Acceleration

Gravitational acceleration, denoted as g, is the acceleration of an object due to Earth's gravity, approximately 9.81 m/s². When the ball is dropped, it accelerates downwards due to gravity, and its velocity upon entering the magnetic field can be calculated using kinematic equations, which is essential for determining the magnetic force.
추천 영상:
가이드 코스
07:32
Weight Force & Gravitational Acceleration
관련 실천
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An open plastic soda bottle with an opening diameter of 2.5 cm is placed on a table. A uniform 1.75 T magnetic field directed upward and oriented 25° from vertical encompasses the bottle. What is the total magnetic flux through the plastic of the soda bottle?

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A horizontal rectangular surface has dimensions 2.80 cm by 3.20 cm and is in a uniform magnetic field that is directed at an angle of 30.0° above the horizontal. What must the magnitude of the magnetic field be to produce a flux of 3.10 x 10-4 Wb through the surface?

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A deuteron (the nucleus of an isotope of hydrogen) has a mass of 3.34 x 10-27 kg and a charge of +e. The deuteron travels in a circular path with a radius of 6.96 mm in a magnetic field with magnitude 2.50 T. (a) Find the speed of the deuteron. (b) Find the time required for it to make half a revolution. (c) Through what potential difference would the deuteron have to be accelerated to acquire this speed?

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A circular area with a radius of 6.50 cm lies in the xy-plane. What is the magnitude of the magnetic flux through this circle due to a uniform magnetic field B = 0.230 T at an angle of 53.1° from the +z-direction?

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