Physics Objective Questions and Answers: Moving Charges and Magnetism
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Multiple Choice Questions
(a) 8.0 A
(b) 4.0 A
(c) 2.0 A
(d) 1.0 A
Show Answer & Explanation
Answer: (a) 8.0 A
Explanation:
1. \( n = 2000 \) turns/m.
2. I = \( B/(\mu_{0} n) = 0.02/(4 \pi x 10^{-7} x 2000) = 7.96 \), about \( 8 A \).
Teacher's Note:
1. \( B = \mu_{0} n \) I.
(a) Biot-Savart's coils
(b) Ampere's coils
(c) Helmholtz coils
(d) Oersted's coils
Show Answer & Explanation
Answer: (c) Helmholtz coils
Explanation:
1. This arrangement produces a nearly uniform field in the region between the coils.
2. It is the Helmholtz coil arrangement.
Teacher's Note:
1. Helmholtz coils: separation = radius.
(a) \( 2\times10^{3}\ \text{N/C} \)
(b) \( 1.5\times10^{3}\ \text{N/C} \)
(c) \( 3\times10^{3}\ \text{N/C} \)
(d) \( \tfrac43\times10^{-3}\ \text{N/C} \)
Show Answer & Explanation
Answer: (c) \( 3\times10^{3}\ \text{N/C} \)
Explanation:
1. For zero deflection \( E = v B = 2 \times 10^{3} x 1.5 = 3 \times 10^{3} N/C \).
Teacher's Note:
1. \( E \) = vB.
(a) electric field
(b) magnetic field
(c) dielectric strength
(d) heat
Show Answer & Explanation
Answer: (b) magnetic field
Explanation:
1. The current produces a magnetic field and the energy is stored in it (\( U = (1/2) L I^{2} \).
Teacher's Note:
1. Energy sits in the \( B \) field.
(a) \( \frac{\mu_0 i}{2}\left[\frac1{R_1}+\frac1{R_2}+\frac1{R_3}+\cdots\right] \)
(b) \( \frac{\mu_0 i}{2}\left[\frac1{R_1}-\frac1{R_2}+\frac1{R_3}-\cdots\right] \)
(c) \( \frac{\mu_0 i}{4\pi}\left[\frac1{R_1}+\frac1{R_2}+\frac1{R_3}+\cdots\right] \)
(d) \( \frac{\mu_0 i}{4\pi}\left[\frac1{R_1}-\frac1{R_2}+\frac1{R_3}-\cdots\right] \)
Show Answer & Explanation
Answer: (b) \( \frac{\mu_0 i}{2}\left[\frac1{R_1}-\frac1{R_2}+\frac1{R_3}-\cdots\right] \)
Explanation:
1. Each loop gives \( B = \mu_{0} i/(2R) \).
2. Neighbouring loops have opposite senses, so their contributions alternate in sign.
Teacher's Note:
1. Alternate loops oppose.
(a) 4B
(b) B/2
(c) B
(d) 2B
Show Answer & Explanation
Answer: (c) B
Explanation:
1. \( B = \mu_{0} n \) I.
2. Doubling I and halving \( n \) leaves the product nI, and hence \( B \), unchanged.
Teacher's Note:
1. \( n \) I is unchanged.
(a) \( \frac{Bq}{mv} \)
(b) \( \frac{mq}{Bv} \)
(c) \( \frac{mB}{qv} \)
(d) \( \frac{mv}{Bq} \)
Show Answer & Explanation
Answer: (d) \( \frac{mv}{Bq} \)
Explanation:
1. The magnetic force provides the centripetal force: qvB = mv²/r.
2. So \( r \) = mv/(Bq).
Teacher's Note:
1. \( r \) = mv/(qB).
(a) \( NA^2B^2I \)
(b) \( NABI^2 \)
(c) \( N^2ABI \)
(d) \( NABI \)
Show Answer & Explanation
Answer: (d) \( NABI \)
Explanation:
1. Torque = \( M B = N \) I \( A B \) in a radial field.
Teacher's Note:
1. Torque = NIAB.
(a) directly proportional to its length
(b) directly proportional to the current
(c) inversely proportional to the total number of turns
(d) inversely proportional to the current
Show Answer & Explanation
Answer: (b) directly proportional to the current
Explanation:
1. \( B = \mu_{0} n i \), where \( n \) is the number of turns per unit length.
2. So \( B \) is directly proportional to the current.
Teacher's Note:
1. \( B = \mu_{0} n i \).
(a) 0.01 N m
(b) 0.001 N m
(c) zero
(d) 0.8 N m
Show Answer & Explanation
Answer: (c) zero
Explanation:
1. The plane is perpendicular to \( B \), so the magnetic moment is parallel to \( B \).
2. Torque = \( M B \sin(0) = 0 \).
Teacher's Note:
1. Torque is zero when \( M \) is parallel to \( B \).
(a) i r B
(b) 2 pi r i B
(c) \( \text{zero} \)
(d) pi r i B
Show Answer & Explanation
Answer: (c) \( \text{zero} \)
Explanation:
1. The forces on diametrically opposite elements are equal and opposite.
2. The net force on the whole loop is zero (it only tends to stretch the loop).
Teacher's Note:
1. Closed loop: zero net force.
(a) directly proportional to the torsional constant
(b) directly proportional to the number of turns
(c) inversely proportional to the area of the coil
(d) inversely proportional to the current flowing
Show Answer & Explanation
Answer: (b) directly proportional to the number of turns
Explanation:
1. \( \theta = N \) I \( A B / k \).
2. Deflection is proportional to \( N \), I, \( A \) and \( B \), and inversely proportional to the torsional constant \( k \).
Teacher's Note:
1. θ = NIAB/k.
(a) 30 degrees
(b) 45 degrees
(c) 60 degrees
(d) 90 degrees
Show Answer & Explanation
Answer: (a) 30 degrees
Explanation:
1. \( F = B \) I \( l \sin(\theta) \), so \( \sin(\theta) = 15/(2 \times 10 \times 1.5) = 0.5 \).
2. Hence \( \theta = 30^{\circ} \).
Teacher's Note:
1. \( \sin(\theta) \) = F/(BIl).
(a) f is proportional to i, a straight line through the origin
(b) f is proportional to the square root of i
(c) f is proportional to i squared, a parabola through the origin
(d) f is inversely proportional to i
Show Answer & Explanation
Answer: (c) f is proportional to i squared, a parabola through the origin
Explanation:
1. \( f = \mu_{0} i1 i2/(2 \pi d) = \mu_{0} i^{2}/(2 \pi d) \).
2. So \( f \) is proportional to \( i^{2} \).
Teacher's Note:
1. \( f \) is proportional to \( i \) squared.
(a) directly proportional to r
(b) inversely proportional to I
(c) directly proportional to I
(d) directly proportional to I squared
Show Answer & Explanation
Answer: (c) directly proportional to I
Explanation:
1. \( B = \mu_{0} I/(2r) \).
2. So \( B \) is proportional to I and inversely proportional to \( r \).
Teacher's Note:
1. \( B = \mu_{0} I/2r \).
(a) The steady current which, maintained in each of two very long, straight, parallel conductors of negligible cross-section placed one metre apart in vacuum, produces a force of \( 2\times10^{7} \) N per metre of length on each
(b) The steady current which, maintained in two such conductors placed one centimetre apart in vacuum, produces a force of \( 2\times10^{-7} \) N per metre of length on each
(c) The steady current which, maintained in two such conductors placed one metre apart in vacuum, produces a force of \( 2\times10^{-7} \) N per metre of length on each
(d) The steady current which, maintained in two such conductors placed ten metres apart in vacuum, produces a force of \( 2\times10^{-7} \) N per metre of length on each
Show Answer & Explanation
Answer: (c) The steady current which, maintained in two such conductors placed one metre apart in vacuum, produces a force of \( 2\times10^{-7} \) N per metre of length on each
Explanation:
1. \( F/L = \mu_{0} I1 I2/(2 \pi d) = 2 \times 10^{-7} N/m \) for \( I1 = I2 = 1 A \) and \( d = 1 m \).
Teacher's Note:
1. \( 1 A, 1 m \) gives \( 2 \times 10^{-7} N/m \).
(a) \( \vec v=\vec B\times\vec E/E^2 \)
(b) \( \vec v=\vec E\times\vec B/B^2 \)
(c) \( \vec v=\vec B\times\vec E/B^2 \)
(d) \( \vec v=\vec E\times\vec B/E^2 \)
Show Answer & Explanation
Answer: (b) \( \vec v=\vec E\times\vec B/B^2 \)
Explanation:
1. The electric and magnetic forces must cancel: qE \( + q(v x B) = 0 \), i.e. \( E = -(v x B) \).
2. Taking the cross product with \( B \) gives \( v = E x B/B^{2} \).
Teacher's Note:
1. Forces balance in a velocity selector.
(a) \( \mu_0 ni \)
(b) \( \frac{\mu_0 ni}{2\pi r} \)
(c) zero
(d) \( \frac{\mu_0 ni}{4\pi r} \)
Show Answer & Explanation
Answer: (c) zero
Explanation:
1. The wires are placed symmetrically, so opposite wires produce equal and opposite fields on the axis.
2. The resultant is zero (also by Ampere's law, since no current is enclosed by a circle of small radius).
Teacher's Note:
1. Symmetry cancels the fields.
(a) the velocity remains unchanged
(b) the speed of the particle remains unchanged
(c) the direction of the particle remains unchanged
(d) the acceleration remains unchanged
Show Answer & Explanation
Answer: (b) the speed of the particle remains unchanged
Explanation:
1. The force is perpendicular to \( v \), so it changes only the direction of \( v \).
2. The speed stays constant.
Teacher's Note:
1. Speed is constant, direction changes.
Assertion (A): If a charged particle is released from rest in a region of uniform electric and magnetic fields parallel to each other, it will move in a straight line.
Reason (R): The electric field exerts no force on the particle but the magnetic field does.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (c) A is true, but R is false.
Explanation:
1. The electric force acts along the common field direction and the particle starts moving along it.
2. The magnetic force is zero for motion parallel to \( B \), so the path is a straight line.
3. The reason is false because it is the electric field that exerts the force.
Teacher's Note:
1. Magnetic force is zero when \( v \) is parallel to \( B \).
(a) \( 3\times10^{-4}\ \text{N} \)
(b) \( 3\times10^{-2}\ \text{N} \)
(c) \( 3\times10^{2}\ \text{N} \)
(d) \( 3\times10^{4}\ \text{N} \)
Show Answer & Explanation
Answer: (b) \( 3\times10^{-2}\ \text{N} \)
Explanation:
1. \( B = 500 \) gauss = \( 0.05 T \).
2. \( F = B \) I \( l \sin(30) = 0.05 \times 3 \times 0.4 \times 0.5 = 3 \times 10^{-2} N \).
Teacher's Note:
1. 1 gauss = \( 1e-4 T \).
(a) \( d\vec B=\frac{\mu_0}{4\pi}i\left(\frac{d\vec\ell\times\vec r}{r}\right) \)
(b) \( d\vec B=\frac{\mu_0}{4\pi}i^2\left(\frac{d\vec\ell\times\vec r}{r}\right) \)
(c) \( d\vec B=\frac{\mu_0}{4\pi}i^2\left(\frac{d\vec\ell\times\vec r}{r^2}\right) \)
(d) \( d\vec B=\frac{\mu_0}{4\pi}i\left(\frac{d\vec\ell\times\vec r}{r^3}\right) \)
Show Answer & Explanation
Answer: (d) \( d\vec B=\frac{\mu_0}{4\pi}i\left(\frac{d\vec\ell\times\vec r}{r^3}\right) \)
Explanation:
1. The Biot-Savart law gives dB = \( (\mu_{0}/4pi) i \) dl \( \sin(\theta)/r^{2} \).
2. In vector form this is \( (\mu_{0}/4pi) i \) (dl x r)/r³.
Teacher's Note:
1. dB follows the inverse square law.
(a) 0.495 ohm
(b) 49.5 ohm
(c) 495 ohm
(d) 4950 ohm
Show Answer & Explanation
Answer: (c) 495 ohm
Explanation:
1. Full-scale current = 10 mV/5 ohm = 2 mA.
2. Total resistance needed = \( 1 V/2 \) mA = 500 ohm, so the series resistance is \( 500 - 5 = 495 \) ohm.
Teacher's Note:
1. \( R \) = V/Ig \( - G \).
(a) \( \mu_0/17 \)
(b) \( \sqrt3\,\mu_0/2\pi \)
(c) \( \mu_0/2\pi \)
(d) \( 3\mu_0/2\pi \)
Show Answer & Explanation
Answer: (c) \( \mu_0/2\pi \)
Explanation:
1. The point is \( 2.5 m \) from each wire.
2. \( B_P = \mu_{0} \times 2.5/(2 \pi x 2.5) = \mu_{0}/(2 \pi) \) and \( B_Q = \mu_{0} \times 5/(2 \pi x 2.5) = \mu_{0}/\pi \).
3. For currents in the same direction the fields at the mid-point are opposite, so \( B = \mu_{0}/\pi - \mu_{0}/(2 \pi) = \mu_{0}/(2 \pi) \).
Teacher's Note:
1. Fields oppose at the midpoint.
(a) turn towards the right of the direction of motion
(b) decrease its speed
(c) increase its speed
(d) turn towards the left of the direction of motion
Show Answer & Explanation
Answer: (b) decrease its speed
Explanation:
1. The magnetic force is zero because \( v \) is parallel to \( B \).
2. The electric force on the electron is opposite to \( E \), so it decelerates it.
Teacher's Note:
1. Electron force is opposite to \( E \).
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