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If a proton is exposed to an external magnetic field of 2 T that has a direction perpendicular to the axis of the proton's spin, what will be the torque on the proton? (a) 0; (b) 1.4 \(\times\) 10\(^{-26}\) N \(\cdot\) m; (c) 2.8 \(\times\) 10\(^{-26}\) N \(\cdot\) m; (d) 0.7 \(\times\) 10\(^{-26}\) N \(\cdot\) m.

Short Answer

Expert verified
The torque is (c) 2.8 \(\times\) 10\(^{-26}\) N \cdot m.

Step by step solution

01

Understanding the Formula for Torque

The torque (\(\tau\)) on a magnetic moment (\(\mu\)) in a magnetic field (\(B\)) is given by the formula:\[\tau = \mu \times B\]where the direction is given by the right-hand rule. Since the magnetic field is perpendicular to the spin axis of the proton, we can use the magnitude formula:\[\tau = \mu B \sin(\theta)\]with \(\theta = 90^\circ\), and thus\(\sin(\theta) = 1\).
02

Finding the Magnetic Moment of Proton

The magnetic moment (\(\mu\)) of a proton is a known physical constant: \(\mu = 1.41 \times 10^{-26}\) A \(\cdot\) m\(^2\).
03

Calculating the Torque

Substitute the known values into the torque formula:\(\tau = \mu \cdot B \cdot \sin(\theta)\), where \(B = 2\) T and \(\theta = 90^\circ\). Thus:\[\tau = (1.41 \times 10^{-26})(2)(1)\]which simplifies to:\[\tau = 2.82 \times 10^{-26} \text{ N} \cdot \text{m}\]
04

Matching the Calculated Torque with Given Options

Compare the calculated torque,\(2.82 \times 10^{-26}\) N \(\cdot\) m, with the options provided. It matches closely with option (c) \(2.8 \times 10^{-26}\) N \(\cdot\) m.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Magnetic Moment
The concept of magnetic moment is essential in understanding the behavior of protons under external magnetic influences. A magnetic moment (\(\mu\)) is a vector quantity that represents the source of a magnetic field and its strength. For a proton, the magnetic moment can be thought of as resulting from its spin and charge. It's a tiny magnet within the nucleus that interacts with external magnetic fields.

When a proton is placed in a magnetic field, its magnetic moment tends to align with the field. The magnetic moment of a proton is approximately \(1.41 \times 10^{-26}\) A \(\cdot\) mtwosup>2. This constant provides a measure of how the proton's intrinsic properties affect its interaction with magnetic fields.
  • Magnitude of magnetic moment tells us the strength of the magnetic effect due to proton's spin.
  • Direction denotes the alignment of the proton in a magnetic field.
External Magnetic Field
Magnetic fields are invisible forces that exert influence on moving charges. An external magnetic field is any magnetic field applied to a region or object from outside its own magnetism. In this context, we consider an external magnetic field of 2 Tesla, which is a measure of the field's strength.

The direction of the magnetic field relative to the proton's magnetic moment is crucial. If the field is perpendicular to the spin of the proton, maximum torque is produced, causing significant alignment of the proton's magnetic moment.
  • Fields of high strength can enact greater forces on small magnetic moments.
  • A perpendicular magnetic field exerts the highest possible torque.
Right-Hand Rule
The right-hand rule is a convenient way to determine the direction of a vector resulting from a cross-product operation in physics. For torque, it is often used to find the torque direction when a magnetic moment is subjected to a magnetic field.

To apply the right-hand rule, point your right thumb in the direction of the proton's magnetic moment (\(\mu\)), and your fingers in the direction of the magnetic field (\(B\)). The direction that your palm pushes represents the direction of the torque (\(\tau\)). This rule ensures that one can determine the rotational effect on the proton without visualizing the actual vectors.
  • Thumb points in the direction of magnetic moment.
  • Fingers point along the direction of the magnetic field.
  • Palm or resultant force indicates the direction of torque.
This simple mnemonic aids in visualizing complex magnetic interactions effectively.

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Most popular questions from this chapter

A particle with charge 7.80 \(\mu\)C is moving with velocity \(\vec{v} =\) - 13.80 \(\times\) 103m/s\(\hat{\jmath}\). The magnetic force on the particle is measured to be \(\overrightarrow{F} =\) (7.60 \(\times\) 10\(^{-3}\) N)\(\hat{\imath}\) - (5.20 \(\times\) 10\(^{-3}\) N)\(\hat{k}\). (a) Calculate all the components of the magnetic field you can from this information. (b) Are there components of the magnetic field that are not determined by the measurement of the force? Explain. (c) Calculate the scalar product \(\overrightarrow{B}\) \(\cdot\) \(\overrightarrow{F}\). What is the angle between \(\overrightarrow{B}\) and \(\overrightarrow{F}\)?

A proton (\(q\) = 1.60 \(\times\) 10\(^{-19}\) C, \(m =\) 1.67 \(\times\) 10\(^{-27}\) kg) moves in a uniform magnetic field \(\overrightarrow{B} =\) (0.500 T)\(\hat{\imath}\). At \(t =\) 0 the proton has velocity components \(\upsilon_x =\) 1.50 \(\times\) 10\(^5\) m/s, \(\upsilon_y =\) 0, and \(\upsilon_z =\) 2.00 \(\times\) 10\(^5\) m/s (see Example 27.4). (a) What are the magnitude and direction of the magnetic force acting on the proton? In addition to the magnetic field there is a uniform electric field in the +\(x\)-direction, \(\overrightarrow{E} =\) (+2.00 \(\times\) 10\(^4\) V/m)\(\hat{\imath}\). (b) Will the proton have a component of acceleration in the direction of the electric field? (c) Describe the path of the proton. Does the electric field affect the radius of the helix? Explain. (d) At \(t =\) \(T\)/2, where T is the period of the circular motion of the proton, what is the \(x\)-component of the displacement of the proton from its position at \(t =\) 0?

An electron experiences a magnetic force of magnitude 4.60 \(\times\) 10\(^{-15}\) N when moving at an angle of 60.0\(^\circ\) with respect to a magnetic field of magnitude 3.50 \(\times\) 10\(^{-3}\) T. Find the speed of the electron.

A particle with initial velocity \(\vec{v}$$_0 =\) (5.85 \(\times\) 10\(^3\)m/s)\(\hat{\jmath}\) enters a region of uniform electric and magnetic fields. The magnetic field in the region is \(\overrightarrow{B} =\) - (1.35 T)\(\hat{k}\). Calculate the magnitude and direction of the electric field in the region if the particle is to pass through undeflected, for a particle of charge (a) +0.640 nC and (b) -0.320 nC. You can ignore the weight of the particle.

A deuteron (the nucleus of an isotope of hydrogen) has a mass of 3.34 \(\times\) 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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