Magnetic Effects Of Current Test

Result

Magnetic Effects Of Current Test - 6

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Question 1
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A uniform magnetic field B is directed out of the page. A metallic wire has the shape of a square frame and is placed in the field as shown. While the shape of the wire is steadily transformed into a circle in the same plane, the current in the frame:

A
is directed clockwise

B
does not appear

C
is directed counterclockwise

D
is alternating

Solution

So, current flows clockwise (by lenz law).
Question 2
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The current I in the straight conductor varies sinusoidally with time as shown. (A positive value of I is in the direction indicated). At time t₁ the induced current in the rectangular loop is clockwise. What is the current in the rectangular loop at time t₂and time t₃ respectively?

A
zero; clockwise

B
zero; counterclockwise

C
clockwise; clockwise

D
clockwise; counterclockwise

Solution

For t₂ : : I is continuously decreasing, so magnetic field is downward direction is decreasing, hence direction of induced current will be clockwise. For t₃ I is continuously increasing in downward direction, so magnetic field in direction outside the plane of paper is increasing. So, again clockwise.
Question 3
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A charge +2q moves vertically upwards with speed v, a second charge − q moves horizontally to the right with the same speed v, and a third charge +q moves horizontally to the right with the same speed v. The point P is located a perpendicular distance a away from each charge as shown in the figure. The magnetic field at point P is

A
Into the page with magnitude

B
Into the page with magnitude

C
Into the page with magnitude

D
Out the page with magnitude

Solution
Question 4
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For the solenoids shown in the diagram (which are assumed to be close to each other), the resistance of the left-hand circuit is slowly increased. In which direction does the current flow through galvanometer in the right-hand circuit?

A
B to A

B
A to B

C
there is no current

D
cannot be determined

Solution

Increasing the resistance causes a decrease in the current on the left. This reduces the field strength for the solenoid B = µ₀nI. By the right-hand rule, the field through the solenoid is directed to the RIGHT. Hence, in the right-hand circuit, since there will be fewer field lines directed to the right, there is an induced electric field that will produce a current in the wires to "replace" the lost field lines. This means that current will be directed with the same orientation as the current in the left circuit... the galvanometer will deflect to the LEFT.
Question 5
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The wire shown in figure carries a current of 60A. Determine the magnitude of the magnetic field at the centre O. Given radius of the bent coil is 2 cm.

A
1.414 × 10^-3 T

B
1.414 × 10^-4 T

C
1.414 × 10^-6 T

D
1.414 × 10^-5 T

Solution

According to Biot-Savart law, magnetic field at the centre O is
Question 6
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A circular loop of radius R = 20 cm is placed in a uniform magnetic field in x-y plane as shown. The loop carries a current i = 1A in the direction as shown in the figure. The torque acting on the loop is

A
1N – m

B
0.25 N – m

C
2 N – m

D
None of these

Solution
Question 7
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Shown in the figure is a conductor carrying a current I. The magnetic field intensity at the point O(common centre of all the three arcs) is :

A

B

C

D

Solution

Hint: Find B due to each arc at O and add them using vector addition.

Since, magnetic field at the centre of an arc is equal to
Question 8
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Consider a toroid of circular cross-section of radius b, major radius R much greater than minor radius b, (see diagram) find the total energy stored in magnetic field of toroid –

A

B

C

D

Solution
Question 9
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Two circular coils made of similar wires but of radius 20 cm and 40 cm are connected in parallel. The ratio of magnetic fields at their centre is -

A
4: 1

B
1: 4

C
2: 1

D
1: 2

Solution
Question 10
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Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross-sectional area A = 10 cm² and length l = 20 cm. If one of the solenoids has 300 turns and the other 400 turns, their mutual-inductance is

A
2.4 π × 10^-5 H

B
4.8 π × 10^-4 H

C
4.8 π × 10^-5 H

D
2.4 π × 10^-5 H

Solution